Suppose \(a, b, c, d\) are real numbers with \(a < b\) and \(c < d.\) -By the countable subadditivity of the outer measure
+By thecountable subadditivity of the outer measure
\[|(a, b) \cup (c, d)| \leq (b - a) + (d - c)\]
always holds. @@ -1323,79 +1323,6 @@
Chapter 2.APutting these results together arrive at the required conclusion.
-
-Excercise 2.A.8
-Prove that if \(A \subseteq \mathbb{R}\) and \(t > 0,\) then \(|A| = |A \cap (-t, t)| + |A \cap (\mathbb{R} \setminus (-t, t))|.\)
-
-
-Solution
-Solution
-
-
-
-
-
-First, by the countable subadditivity of the outer measure we have
-
-\[|A| \leq |A \cap (-t, t)| + |A \cap (\mathbb{R} \setminus (-t, t))|\]
-for all \(t > 0.\) -To prove the inequality the other way, suppose \(I_1, I_2, \dots\) is a sequence of open intervals whose union contains \(A.\) -Then, we have
-
-\[\begin{split}\begin{align}
-\sum_{n = 1}^\infty \ell(I_n) &= \sum_{n = 1}^\infty \ell(I_n \cap (-t, t)) + \ell(I_n \cap (-\infty, t)) + \ell(I_n \cap (\mathbb{R} \setminus (t, \infty))) \\
-&\geq |A \cap (-t, t)| + |A \cap (\mathbb{R} \setminus (-t, t))|
-\end{align}\end{split}\]
-where we have used the fact that the sequence of sets
-
-\[I_1 \cap (-\infty, t], I_1 \cap [t, \infty), I_2 \cap (-\infty, t], \dots\]
-has a union that contains \(A \cap (\mathbb{R} \setminus (-t, t)),\) and the outer measures of these sets are equal to
-
-\[\ell(I_1 \cap (-\infty, t)), \ell(I_1 \cap (t, \infty)), \ell(I_2 \cap (-\infty, t)), \dots,\]
-completing the proof.
-
-
-Exercise 2.A.9
-Prove that \(|A| = \lim_{t \to \infty} |A \cap (-t, t)|\) for all \(A \subseteq \mathbb{R}.\)
-
-
-Solution
-Solution
-
-
-
-
-
-First, by the countable subadditivity of the outer measure we have
-
-\[\begin{split}\begin{align}
-|A| &= \left|\bigcup_{n = 1}^\infty (A \cap (n-1, n)) \cup (A \cap (-n, -n + 1)) \cup \{-n-1, n-1\} \right| \\
-&\leq \sum_{n = 1}^\infty \left|(A \cap (n-1, n)) \cup (A \cap (-n, -n+1)) \cup \{-n-1, n-1\} \right| \\
-&= \lim_{N \to \infty} \sum_{n = 1}^N \left|(A \cap (n-1, n)) \cup (A \cap (-n, -n+1)) \cup \{-n-1, n-1\} \right| \\
-&= \lim_{N \to \infty} \left|A \cap (-N, N)\right|.
-\end{align}\end{split}\]
-Note that since \(\left|A \cap (-t, t)\right|\) is non-decreasing in \(t \in \mathbb{Z}^+,\) the limit above is unchanged even if \(N \not \in \mathbb{Z}^+.\) -In addition, we have \(|A| \geq |A \cap (-t, t)|\) for all \(t \in \mathbb{R},\) and putting these two inequalities together, we conclude that
-
-\[|A| = \lim_{t \to \infty} |A \cap (-t, t)|.\]
-
-
Exercise 2.A.10
-Prove that \(|[0, 1] \setminus \mathbb{Q}| = 1.\)
-
-
-Solution
-Solution
-
-
-
-
-
-First, by the countable subadditivity of the outer measure we have
-
-\[\begin{equation}
-|[0, 1] \setminus \mathbb{Q}| \geq |[0, 1]| - |\mathbb{Q}| = |[0, 1]| = 1,
-\end{equation}\]
-where we have used the fact that countable sets have measure zero. -Using the fact that the outer measure preserves order we have \(|[0, 1] \setminus \mathbb{Q}| \leq |[0, 1]| = 1,\) concluding the proof.
-Chapter 2.C#
@@ -1664,7 +1591,7 @@Chapter 2.C
Exercise 2.C.12
-Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countable or \(X \setminus E\) is countable.
+
Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countatble or \(X \setminus E\) is countable.
Give a complete description of the set of all measures \(\mu\) on \((X, S).\)
@@ -1673,7 +1600,7 @@ Chapter 2.C
-Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countable or \(X \setminus E\) is countable.
+
Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countatble or \(X \setminus E\) is countable.
Then, a measure \(\mu\) on \((X, S)\) is completely determined by the value of \(\mu(\{x\})\) for each \(x \in X,\) along with \(\mu(X).\)
@@ -2080,7 +2007,7 @@ Chapter 2.D
\[|A \cap (n, n+1)| \leq \left| \bigcup_{i=1}^\infty (r_i + V) \right| \leq \sum_{i=1}^\infty |(r_i + [n, n+1])| = 3.\]
Since \(V\) is Lebesgue measurable, so is \(r_i + V\) for each \(i \in \mathbb{N}.\)
-Thus by the countable subadditivity of the outer measure, we have
+Thus by the countable subadditivity of the outer measure, we have
\[ 3 \geq \left| \bigcup_{i=1}^\infty (r_i + V) \right| = \sum_{i=1}^n |r_i + V| \geq \sum_{i=1}^n |V| = n |V|\]
for all \(n \in \mathbb{N},\) which can only hold if \(|V| = 0.\)
@@ -2350,498 +2277,6 @@
Chapter 5.A
-Chapter 6.A#
-
-Exercise 6.A.1
-Verify that each of the following examples of sets \(V\) and functions \(d: V \times V \to \mathbb{R}\) are indeed metric spaces.
-
-Suppose \(V\) is a nonempty set.
-Define \(d\) as
-
-\[\begin{split}\begin{equation}
-d(f, g) = \begin{cases}
-1 &\text{ if } f = g \\
-0 &\text{ if } f \neq g
-\end{cases}
-\end{equation}\end{split}\]
-
-Let \(V = \mathbb{R}.\)
-Define \(d\) as
-
-\[\begin{equation}
-d(x, y) = |x - y|.
-\end{equation}\]
-
-Let \(V = \mathbb{R}.\)
-For \(n \in \mathbb{Z}^+,\) define \(d\) as
-
-\[\begin{equation}
-d((x_1, \dots, x_n), (y_1, \dots, y_n)) = \max\{\|x_1 - y_1|, \dots, |x_n - y_n|\}.
-\end{equation}\]
-
-Let \(V = C([0, 1]),\) the set of continuous real-valued functions on \([0, 1].\)
-Define \(d\) as
-
-\[\begin{equation}
-d(f, g) = \sup\{|f(t) - g(t)|: t \in [0, 1]|\}.
-\end{equation}\]
-
-Let \(V\) be the set of sequences \((a_1, a_2, \dots)\) with \(a_k \in \mathbb{R}\) and \(\sum_{n=1}^\infty |a_k| < \infty.\)
-Define \(d\) as
-
-\[\begin{equation}
-d((a_1, a_2, \dots), (b_1, b_2, \dots)) = \sum_{n=1}^\infty |a_k - b_k|.
-\end{equation}\]
-
-
-Solution
-
-
-
-
-For all the definitions above, \(d(f, g) \geq 0\) and equality with zero holds only if \(f, g.\)
-Further, \(d(f, g) = d(g, f).\)
-It remains to show the triangle inequality
-
-\[\begin{equation}
-d(f, h) \leq d(f, g) + d(g, h) \text{ for all } f, g, h \in V
-\end{equation}\]
-holds for each example.
-Example 1:
-Suppose \(f, g, h \in \mathbb{V}.\)
-If \(f = h,\) then the triangle inequality holds since
-
-\[\begin{equation}
-d(f, h) = 0 \leq d(f, g) + d(g, h),
-\end{equation}\]
-holds.
-If \(f \neq h,\) then at least one of \(f \neq g\) and \(g \neq h\) must hold, so the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = 1 \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 2:
-Suppose \(f, g, h \in \mathbb{R}.\)
-Then the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = |f - h| = |(f - g) + (g - h)| \leq |f - g| + |g - h| \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 3:
-Suppose \(n \in \mathbb{Z}^+\) and \((x_1, \dots, x_n), (y_1, \dots, y_n), (z_1, \dots, z_n) \in \mathbb{R}^n.\)
-Then the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((x_1, \dots, x_n), (z_1, \dots, z_n)) = \\
-&= \max\{|x_1 - z_1|, \dots, |x_n - z_n|\} \\
-&= \max\{|(x_1 - y_1) + (y_1 - z_1)|, \dots, |(x_n - y_n) + (y_n - z_n)|\} \\
-&\leq \max\{|x_1 - y_1| + |y_1 - z_1|, \dots, |x_n - y_n| + |y_n - z_n|\} \\
-&\leq \max\{|x_1 - y_1|, \dots, |x_n - y_n|\} + \max\{|y_1 - z_1|, \dots, |y_n - z_n|\} \\
-&\leq d((x_1, \dots, x_n), (y_1, \dots, y_n)) + d((y_1, \dots, y_n), (z_1, \dots, z_n)).
-\end{align}\end{split}\]
-Example 4:
-Suppose \(f, g, h \in C([0, 1]).\)
-Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-d(f, h) &= \sup\{|f(t) - h(t)|: t \in [0, 1]|\} \\
-&= \sup\{|(f(t) - g(t)) + (g(t) - h(t))|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)| + |g(t) - h(t)|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)|: t \in [0, 1]|\} + \sup\{|g(t) - h(t)|: t \in [0, 1]|\} \\
-&= d(f, g) + d(g, h).
-\end{align}\end{split}\]
-Example 5:
-Suppose \((f_1, f_2, \dots), (g_1, g_2, \dots), (h_1, h_2, \dots) \in V,\) where \(V\) is the set of sequences \((a_1, a_2, \dots)\) of real numbers such that \(\sum_{n=1}^\infty |a_k| < \infty.\)
-Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((f_1, f_2, \dots), (h_1, h_2, \dots)) = \\
-&= \sum_{n=1}^\infty |f_k - h_k| \\
-&= \sum_{n=1}^\infty |(f_k - g_k) + (g_k - h_k)| \\
-&\leq \sum_{n=1}^\infty |f_k - g_k| + |g_k - h_k| \\
-&= d((f_1, f_2, \dots), (g_1, g_2, \dots)) + d((g_1, g_2, \dots), (h_1, h_2, \dots)).
-\end{align}\end{split}\]
-
-
-
-Exercise 6.A.2
-Prove that every finite subset of a metric space is closed.
-
-
-Solution
-
-
-
-
-Suppose \((V, d)\) is a metric space and let \(A\) be a finite subset of \(V.\)
-Let \(x \in A'.\)
-Since \(A\) is finite, the set \(\{d(x, y) \in \mathbb{R}^+: y \in A\}\) has a minimum element, and this minimum element is non-zero, say equal to some positive \(m \in \mathbb{R}\)
-Therefore, any open ball centered on \(x\) with radius less than or equal to \(m\) is contained in \(A'.\)
-Since \(x\) was arbitrary, \(A'\) is open, so \(A\) is closed.
-
-
-
-Exercise 6.A.3
-Prove that every closed ball in a metric space is closed.
-
-
-Solution
-
-
-
-
-Suppose \((V, d)\) is a metric space.
-Let \(r \in \mathbb{R}^+\) and \(f \in V.\)
-Let \(g \in \overline{B}(f, r)'.\)
-Then, \(d(f, g) > r\) so the open ball \(B(g, d(f, g) - r)\) does not intersect the closed ball \(\overline{B}(f, r).\)
-[Otherwise there would exist a common element \(h \in B(g, d(f, g) - r) \cap \overline{B}(f, r)\) which leads to a contradiction via the triangle inequality since \(d(f, g) \leq d(f, h) + d(h, g) < r.\)]
-Since the two balls do not intersect, \(B(g, r - d(f, g)) \subseteq \overline{B}(f, r)',\) which means that \(\overline{B}(f, r)'\) is an open set, so \(\overline{B}(f, r)\) is a closed set.
-
-
-
-Exercise 6.A.4
-Suppose \(V\) is a metric space.
-
-Prove that the union of each collection of open subsets of \(V\) is an open subset of \(V.\)
Prove that the intersection of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-
-
-
-
-In the following parts, \(V\) is a metric space and \(\mathcal{A}\) is a collection of subsets of \(V.\)
-Part 1:
-Let \(S = \cup_{A \in \mathcal{A}} A.\)
-If \(s \in S,\) there exists some \(A \in \mathcal{A}\) such that \(s \in A.\)
-Since every \(A \in \mathcal{A}\) is open, there exists an open ball centered on \(s\) that is contained in \(A.\)
-This open ball is also contained in \(S,\) so \(S\) is open.
-Part 2:
-Now suppose that, in addition, \(\mathcal{A} = \{A_1, \dots A_N\}\) is finite.
-If \(s \in S,\) then \(s \in A_n\) for \(n = 1, \dots, N.\)
-Since each \(A\) is open, for each \(A_n \in \mathcal{A},\) there exists an open ball centered on \(s\) with radius \(r_n,\) which is contained in \(A_n.\)
-Now, letting \(r = \min\{r_1, \dots, r_N\}\) we see that the open ball centered on \(s\) with radius \(r\) is contained in each \(A_n\) and therefore it is also contained in their intersection, i.e. it is contained in \(S.\)
-Therefore there exists an open ball centered on \(s \in S,\) which implies that \(S\) is open.
-
-
-
-Exercise 6.A.5
-Suppose \(V\) is a metric space.
-
-Prove that the intersection of each collection of closed subsets of \(V\) is an open subset of \(V.\)
Prove that the union of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-
-
-
-
-The complement of an intersection of a collection of sets is equal to the union of the complements of the sets in the collection.
-Similarly, the complement of a union of a collection of sets is equal to the intersection of the complements of the sets in the collection.
-Therefore, applying the result of the previous exercise we arrive at the two required results.
-
-
-
-Exercise 6.A.6
-
-Prove that if \(V\) is a metric space, \(f \in V,\) and \(r > 0,\) then \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
Give an example of a metric space \(V,\) \(f \in V,\) and \(r > 0\) such that \(\overline{B(f, r)} \neq \overline{B}(f, r).\)
-
-
-Solution
-
-
-
-
-Part 1:
-First, note that \(B(f, r) \subseteq \overline{B}(f, r).\)
-Second \(\overline{B}(f, r)\) is a closed set, and the closure of a set is the intersection of all closed sets that contain it, so \(\overline{B(f, r)}\) is contained in any closed set that contains \(B(f, r),\) so \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
-Part 2:
-Let \(V = \mathbb{Z}\) with the metric \(d: \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}^+\) defined as \(d(f, g) = |f - g|.\)
-Also let \(f = 0\) and \(r = 1.\)
-Then \(B(f, r) = \{0\}\) and so \(\overline{B(f, r)} = \{0\}.\)
-However, \(\overline{B}(f, r) = \{-1, 0, 1\}\) so \(\overline{B(f, r)} \neq \overline{B}(f, r)\) as required.
-
-
-
-Exercise 6.A.7
-Show that each sequence in a metric space has at most one limit.
-
-
-Solution
-
-
-
-
-Let \((V, d)\) be a metric space.
-Suppose \(f_1, f_2, \dots \in V\) is a sequence in \(V.\)
-If \(a, b \in V\) are limits of \(f_1, f_2, \dots,\) then
-
-\[\begin{equation}
-\lim_{n \to \infty} d(f_n, a) = 0 \text{ and } \lim_{n \to \infty} d(f_n, b) = 0.
-\end{equation}\]
-From the triangle inequality, \(d(a, b) \leq d(a, f_n) + d(f_n, b),\) and taking limits of both sides, we conclude that \(d(a, b) = 0,\) which implies that \(a = b.\)
-Therefore the sequence can have at most one limit in \(V.\)
-
-
-
-Exercise 6.A.8
-Prove that each open subset of a metric space \(V\) is the union of some sequence of closed subsets of \(V.\)
-
-
-Solution
-
-
-
-
-Let \(V\) be a metric space, let \(U\) be an open subset of \(V\) and define
-
-\[\begin{equation}
-U_r = \{f \in V: d(f, g) \geq r \text{ for all } g \in U'\}.
-\end{equation}\]
-Therefore \(U_r\) is the set of elements in \(U\) that are at least a distance \(r\) away from \(V.\)
-Note also that \(U_r \subseteq U.\)
-Now, note that for fixed \(g \in U',\) the set \(\{f \in V: d(f, g) \geq r\}\) is closed, because it is the complement of the open set \(\{f \in V: d(f, g) < r\}.\)
-The intersection of a collection of closed sets is closed, so \(U_r\) is closed.
-Now, note that for each \(n \in \mathbb{Z}^+,\) we have \(U_{\frac{1}{n}} \subseteq U,\) from which it follows that \(\bigcup_{n = 1}^\infty U_{\frac{1}{n}} \subseteq U.\)
-Conversely, if \(x \in U,\) since \(U\) is open there exists a ball of radius \(\frac{1}{n}\) for some \(n \in \mathbb{Z}^+\) contained in \(U,\) so \(x \in U_{\frac{1}{n}},\) from which it follows that \(U \subseteq \bigcup_{n = 1}^\infty U_{\frac{1}{n}}.\)
-Therefore \(U\) is the union of a sequence of closed sets in \(V,\) as required.
-
-
-
-Exercise 6.A.10
-Prove or give a counterexample:
-If \(V\) is a metric space and \(U, W\) are subserts of \(V,\) then \(\overline{U} \cup \overline{W} = \overline{U \cup W}.\)
-
-
-Solution
-
-
-
-
-If \(v \in \overline{U},\) then there exists a sequence of elements in \(U\) whose limit is \(v.\)
-Therefore there exists a sequence of elements in \(U \cup W\) whose limit is \(v,\) so \(v \in \overline{U \cup W}.\)
-Similarly, if \(v \in \overline{W},\) it follows that \(v \in \overline{U \cup W}.\)
-We conclude that \(\overline{U} \cup \overline{W} \subseteq \overline{U \cup W}.\)
-If \(v \in \overline{U \cup W},\) then \(v\) must be the limit of a sequence of elements in \(U \cup W.\)
-This sequence must have a infinite subsequence of elements in at least one of \(U\) or \(W\) with \(v\) as its limit, so \(v \in \overline{U} \cup \overline{W}.\)
-We conclude that \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W},\) which completes the proof.
-
-
-
-Exercise 6.A.11
-Prove or give a counterexample:
-If \(V\) is a metric space and \(U, W\) are subsets of \(V,\) then \(\overline{U} \cap \overline{W} = \overline{U \cap W}.\)
-
-
-Solution
-
-
-
-
-The equation does not hold.
-As a counterexample, consider \(\mathbb{R}\) with the metric
-
-\[d(f, g) = |f - g|,\]
-and let \(U = (-1, 0), W = (0, 1).\)
-Then we have \(\overline{U} = [-1, 0]\) and \(\overline{W} = [0, 1].\)
-Therefore, we have \(\overline{U} \cap \overline{W} = \{0\},\) but \(U \cap W = \emptyset\) so \(\overline{U \cap W} = \emptyset.\)
-
-
-
-Exercise 6.A.12
-Suppose \((U, d_U), (V, d_V)\) and \((W, d_W)\) are metric spaces.
-Suppose also that \(T: U \to V\) and \(S: V \to W\) are continuous functions.
-
-Using the definition of continuity, show that \(S \circ T: U \to W\) is continuous.
Using the equivalence of continuity with the property that the limit of a function is equal to the function of its limit, show that \(S \circ T: U \to W\) is continuous.
Using the equivalence of continuity with the property that the inverse image of an open set under a function is open, show that \(S \circ T: U \to W\) is continuous.
-
-
-Solution
-
-
-
-
-Part 1:
-Let \(v \in V\) and \(\epsilon > 0.\)
-Since \(S\) is continuous, there exists \(\delta_S > 0\) such that \(d_W(S(v), S(v')) < \epsilon\) for all \(v' \in V\) such that \(d_V(v, v') < \delta_S.\)
-Let \(u \in U.\)
-Since \(T\) is continuous, there exists \(\delta_T > 0\) such that \(d_V(T(u), T(u')) < \delta_S\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T.\)
-Letting \(v = T(u)\) and putting together these facts, we see that \(d_W(S \circ T(u), S \circ T(u')) < \epsilon\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T,\) which shows that \(S \circ T\) is continuous.
-Part 2:
-Suppose \(u_1, u_2, \dots\) is a sequence in \(U\) with limit \(u \in U.\)
-Since \(S\) and \(T\) are both continuous
-
-\[\begin{split}\begin{align}
-S \circ T(u) &= S(T(u)) \\
-&= S\left(T\left(\lim_{k \to \infty} u_k\right)\right) \\
-&= S\left(\lim_{k \to \infty} T(u_k)\right) \\
-&= \lim_{k \to \infty} S\left(T(u_k)\right) \\
-&= \lim_{k \to \infty} S \circ T (u_k)
-\end{align}\end{split}\]
-Therefore \(S \circ T\) is continuous.
-Part 3:
-Suppose \(G\) is an open subset in \(W.\)
-Since \(S\) is continuous, \(S^{-1}(G)\) is open in \(V\) and since \(T\) is continuous, \(T^{-1}(S^{-1}(G)) = (S \circ T)^{-1}(G)\) is open in \(U,\) so \(S \circ T\) is continuous.
-
-
-
-Exercise 6.A.14
-Suppose a Cauchy sequence in a metric space has a convergent subsequence.
-Prove that the Cauchy sequence converges.
-
-
-Solution
-
-
-
-
-Let \((V, d)\) be a metric space and \(v_1, v_2, \dots\) be a Cauchy sequence in \(V.\)
-Suppose that \(v_{k_1}, v_{k_2}, \dots\) is a subsequence which converges to \(v \in V.\)
-Let \(\epsilon > 0.\)
-Since the sequence \(v_1, v_2, \dots\) is Cauchy, there exists \(K\) such that for all \(k, k' \geq K\) we have \(d(v_k, v_{k'}) < \frac{\epsilon}{2}.\)
-In addition, by our earlier assumption, there exists \(N\) such that for all \(n \geq N\) we have \(d(v_{k_n}, v) < \frac{\epsilon}{2}.\)
-Then, letting \(L\) be the maximum of \(K\) and \(k_N\) we see that for any \(l \geq L\) it holds that \(d(v_l, v) \leq d(v_l, v_k) + d(v_k, v) < \epsilon,\) concluding the proof.
-
-
-
-Exercise 6.A.16
-Suppose \((U, d)\) is a metric space.
-Let \(W\) denote the set of all Cauchy sequences of elements of \(U.\)
-
-For \((f_1, f_2, \dots)\) and \((g_1, g_2, \dots)\) in \(W,\) define \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots)\) to mean that \(\lim_{k \to \infty} d(f_k, g_k) = 0.\)
-Show that \(\equiv\) is an equivalence relation on \(W.\)
Let \(V\) denote the set of equivalence classses of elements of \(W\) under the equivalence relation above.
-For \((f_1, f_2, \dots) \in W,\) let \((f_1, f_2, \dots)\hat{~}\) denote the equivalence class of \((f_1, f_2, \dots).\)
-Show that the following definition of \(d_V: V \times V \to [0, \infty)\) makes sense and that \(d_V\) is a metric on \(V\)
-
-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k).
-\end{equation}\]
-
-Show that \((V, d_V)\) is a complete metric space.
Show that the map from \(U\) to \(V\) that takes \(f \in U\) to \((f, f, f, \dots)\hat{~}\) preserves distances, meaning that for all \(f, g \in U,\) we have
-
-\[d(f, g) = d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~})\]
-
-Explain why (4) shows that every metric space is a subset of some complete metric space.
-
-
-Solution
-
-
-
-
-Part 1:
-First, we have that \(d(f_k, f_k) = 0\) so \((f_1, f_2, \dots) \equiv (f_1, f_2, \dots).\)
-Second, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) we have
-
-\[\lim_{k \to \infty} d(f_k, g_k) = 0 \implies \lim_{k \to \infty} d(g_k, f_k) = 0,\]
-so it follows that \((g_1, g_2, \dots) \equiv (f_1, f_2, \dots).\)
-Third, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) and \((g_1, g_2, \dots) \equiv (h_1, h_2, \dots),\) we have that
-
-\[\lim_{k \to \infty} d(f_k, h_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) = 0,\]
-which means that \((f_1, f_2, \dots) \equiv (h_1, h_2, \dots).\)
-Therefore \(\equiv\) is an equivalence relation.
-Part 2:
-First, for \(d_V\) to be well-defined, it should not matter which representative elements of \((f_1, f_2, \dots)\hat{~}\) and \((g_1, g_2, \dots)\hat{~}\) we pick in the right hand side of the equation that defines \(d_V.\)
-In particular, suppose \((\tilde{f}_1, \tilde{f}_2, \dots) \in (f_1, f_2, \dots)\hat{~}.\)
-Then
-
-\[\begin{equation}
-\lim_{k \to \infty} d(\tilde{f}_k, g_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(f_k, \tilde{f}_k)) = \lim_{k \to \infty} d(f_k, g_k),
-\end{equation}\]
-so it does not matter which representative element of \((f_1, f_2, \dots)\hat{~}\) we pick when defining \(d_V.\)
-The same holds for \((g_1, g_2, \dots)\hat{~},\) so \(d_V\) is well-defined.
-Second, we show that \(d_V\) is a metric.
-By the definition of \(d_V,\) we have that
-
-\[d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) \geq 0,\]
-with equality holding if and only if \(\lim_{k \to \infty} d(f_k, g_k) = 0,\) which in turn holds if and only if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) which is equivalent to \((f_1, f_2, \dots)\hat{~} = (g_1, g_2, \dots)\hat{~}.\)
-In addition, we have that
-
-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = d_V((g_1, g_2, \dots)\hat{~}, (f_1, f_2, \dots)\hat{~})
-\end{equation}\]
-because \(d(f_k, g_k) = d(g_k, f_k).\)
-Lastly, if \((h_1, h_2, \dots) \in W,\) we have that
-
-\[\begin{split}\begin{align}
-&~~~~d_V((f_1, f_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}) =\\
-&= \lim_{k \to \infty} d(f_k, h_k) \\
-&\leq\lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) \\
-&= d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) + d_V((g_1, g_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}),
-\end{align}\end{split}\]
-so \(d_V\) satisfies the triangle inequality, completing the proof that it is a metric.
-Part 3:
-Suppose that \((v_1, v_2, \dots)\) is a Cauchy sequence in \(V.\)
-We will show that there exists an element \(w \in V\) and \(K \in \mathbb{Z}^+\) such that for all \(k \geq K\) we have \(\lim_{i \to \infty} d((v_k)_i, w_i) = 0.\)
-Since each sequence \(v_k\) is itself Cauchy, there exists \(N_k \in \mathbb{Z}^+\) such that \(N_k \geq k\) and for all \(i, j \geq N_k,\) we have \(d(v_i, v_j) < \frac{1}{k}.\)
-Define the terms of the sequence \(w\) to be \(w_k = (v_k)_{N_k}.\)
-We now show that \(w\) is in fact the limit of \((v_1, v_2, \dots).\)
-Let \(\epsilon > 0.\)
-Since \((v_1, v_2, \dots)\) is Cauchy, there exists \(N_{\epsilon}^{(1)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{\epsilon}^{(1)},\) we have
-
-\[\lim_{k \to \infty} d((v_i)_k, (v_j)_k) < \frac{\epsilon}{3}.\]
-In addition, for fixed \(k \in \mathbb{Z}^+\) since each sequence \(v_k\) is Cauchy, there exists \(N_{k, \epsilon}^{(2)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{k, \epsilon}^{(2)},\) we have
-
-\[\lim_{k \to \infty} d((v_k)_i, (v_k)_j) < \frac{\epsilon}{3}.\]
-Now, we have that
-
-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, w_i) \\
-&= d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_{N_i}) \\
-&\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_j) + d((v_i)_j, (v_i)_{N_i})
-\end{align}\end{split}\]
-where in the first line we have applied the triangle inequality, in the second line we have substituted the definition \(w_i = (v_i)_{N_i}\) and in the third line we have again used the triangle inequality again.
-Now, letting \(i, k \geq N_{\epsilon}^{(1)}\) means that the limit of the second term in the inequality, as \(j \to \infty\), is smaller than \(\epsilon / 3.\)
-In addition, letting \(i, j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)})\) means that the first term in the inequality is smaller than \(\epsilon / 3.\)
-Lastly, by the definition of \(N_i,\) letting \(j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)}, N_i)\) means that the last term in the inequality is smaller than \(\frac{1}{i}.\)
-We therefore obtain
-
-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq \lim_{j \to \infty} d((v_k)_i, (v_k)_j) + \lim_{j \to \infty} d((v_k)_j, (v_i)_j) + \lim_{j \to \infty} d((v_i)_j, (v_i)_{N_i}) \\
-&< \frac{2\epsilon}{3} + \frac{1}{i}
-\end{align}\end{split}\]
-so for all \(i > \frac{3}{\epsilon}\) we have \(d((v_k)_i, w_i) < \epsilon.\)
-This means that \(d_V(v_k\hat{~}, w\hat{~}) < \epsilon\) for all \(k \geq N_{\epsilon}^{(1)}\) which means that \(w\hat{~}\) is the limit of the sequence \((v_1\hat{~}, v_2\hat{~}, \dots),\) so \(V\) is a complete metric space.
-Part 4:
-By part (2), defining \(f_k = f\) and \(g_k = g\) for all \(k \in \mathbb{Z}^+,\) we have that
-
-\[d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k) = d(f, g),\]
-as required.
-Part 5:
-We can add elements to the set \(U\) to ensure that the resulting set is complete.
-Specifically, we add to \(U\) the set
-\(S = \left\{w \in W | w \neq (f, f, f, \dots)\hat{~} \text{ for any } f \in U \right\},\)
-to obtain the larger set \(\hat{U} = U \cup S.\)
-Now, for \(u \in \hat{U},\) we define \(\hat{u}\) to be equal to \((u, u, \dots)\hat{~}\) if \(u \in U\) and equal to \(u\) otherwise.
-Then, define the function \(d_\hat{U}: \hat{U} \times \hat{U} \to \mathbb{R}\) as
-
-\[d_\hat{U}(u, v) = d_V(\hat{u}, \hat{v}).\]
-This function is a metric over \(\hat{U}.\)
-Further, if \((u_1, u_2, \dots)\) is a Cauchy sequence of elements of \(U,\) then \((\hat{u_1}, \hat{u_2}, \dots)\) is a Cauchy sequence of equivalence classes in \(V.\)
-Since \(V\) is complete, the sequence \((\hat{u_1}, \hat{u_2}, \dots)\) has a limit, denoted \(\hat{u},\) in \(V.\)
-Therefore the sequence \((u_1, u_2, \dots)\) converges to \(\hat{u}\) in the metric space \((\hat{U}, d_{hat{U}}).\)
-
-
-
@@ -1673,7 +1600,7 @@ Chapter 2.C
-
Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countable or \(X \setminus E\) is countable. +
Suppose \(X\) is a set and \(S\) is the \(\sigma\)-algebra of all subsets \(E\) of \(X\) such that \(E\) is countatble or \(X \setminus E\) is countable. Then, a measure \(\mu\) on \((X, S)\) is completely determined by the value of \(\mu(\{x\})\) for each \(x \in X,\) along with \(\mu(X).\)
Since \(V\) is Lebesgue measurable, so is \(r_i + V\) for each \(i \in \mathbb{N}.\) -Thus by the countable subadditivity of the outer measure, we have
+Thus by thecountable subadditivity of the outer measure, we have
\[ 3 \geq \left| \bigcup_{i=1}^\infty (r_i + V) \right| = \sum_{i=1}^n |r_i + V| \geq \sum_{i=1}^n |V| = n |V|\]
for all \(n \in \mathbb{N},\) which can only hold if \(|V| = 0.\) @@ -2350,498 +2277,6 @@
Chapter 5.A
-Chapter 6.A#
-
-Exercise 6.A.1
-Verify that each of the following examples of sets \(V\) and functions \(d: V \times V \to \mathbb{R}\) are indeed metric spaces.
-
-Suppose \(V\) is a nonempty set.
-Define \(d\) as
-
-\[\begin{split}\begin{equation}
-d(f, g) = \begin{cases}
-1 &\text{ if } f = g \\
-0 &\text{ if } f \neq g
-\end{cases}
-\end{equation}\end{split}\]
-
-Let \(V = \mathbb{R}.\)
-Define \(d\) as
-
-\[\begin{equation}
-d(x, y) = |x - y|.
-\end{equation}\]
-
-Let \(V = \mathbb{R}.\)
-For \(n \in \mathbb{Z}^+,\) define \(d\) as
-
-\[\begin{equation}
-d((x_1, \dots, x_n), (y_1, \dots, y_n)) = \max\{\|x_1 - y_1|, \dots, |x_n - y_n|\}.
-\end{equation}\]
-
-Let \(V = C([0, 1]),\) the set of continuous real-valued functions on \([0, 1].\)
-Define \(d\) as
-
-\[\begin{equation}
-d(f, g) = \sup\{|f(t) - g(t)|: t \in [0, 1]|\}.
-\end{equation}\]
-
-Let \(V\) be the set of sequences \((a_1, a_2, \dots)\) with \(a_k \in \mathbb{R}\) and \(\sum_{n=1}^\infty |a_k| < \infty.\)
-Define \(d\) as
-
-\[\begin{equation}
-d((a_1, a_2, \dots), (b_1, b_2, \dots)) = \sum_{n=1}^\infty |a_k - b_k|.
-\end{equation}\]
-
-
-Solution
-
-
-
-
-For all the definitions above, \(d(f, g) \geq 0\) and equality with zero holds only if \(f, g.\)
-Further, \(d(f, g) = d(g, f).\)
-It remains to show the triangle inequality
-
-\[\begin{equation}
-d(f, h) \leq d(f, g) + d(g, h) \text{ for all } f, g, h \in V
-\end{equation}\]
-holds for each example.
-Example 1:
-Suppose \(f, g, h \in \mathbb{V}.\)
-If \(f = h,\) then the triangle inequality holds since
-
-\[\begin{equation}
-d(f, h) = 0 \leq d(f, g) + d(g, h),
-\end{equation}\]
-holds.
-If \(f \neq h,\) then at least one of \(f \neq g\) and \(g \neq h\) must hold, so the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = 1 \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 2:
-Suppose \(f, g, h \in \mathbb{R}.\)
-Then the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = |f - h| = |(f - g) + (g - h)| \leq |f - g| + |g - h| \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 3:
-Suppose \(n \in \mathbb{Z}^+\) and \((x_1, \dots, x_n), (y_1, \dots, y_n), (z_1, \dots, z_n) \in \mathbb{R}^n.\)
-Then the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((x_1, \dots, x_n), (z_1, \dots, z_n)) = \\
-&= \max\{|x_1 - z_1|, \dots, |x_n - z_n|\} \\
-&= \max\{|(x_1 - y_1) + (y_1 - z_1)|, \dots, |(x_n - y_n) + (y_n - z_n)|\} \\
-&\leq \max\{|x_1 - y_1| + |y_1 - z_1|, \dots, |x_n - y_n| + |y_n - z_n|\} \\
-&\leq \max\{|x_1 - y_1|, \dots, |x_n - y_n|\} + \max\{|y_1 - z_1|, \dots, |y_n - z_n|\} \\
-&\leq d((x_1, \dots, x_n), (y_1, \dots, y_n)) + d((y_1, \dots, y_n), (z_1, \dots, z_n)).
-\end{align}\end{split}\]
-Example 4:
-Suppose \(f, g, h \in C([0, 1]).\)
-Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-d(f, h) &= \sup\{|f(t) - h(t)|: t \in [0, 1]|\} \\
-&= \sup\{|(f(t) - g(t)) + (g(t) - h(t))|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)| + |g(t) - h(t)|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)|: t \in [0, 1]|\} + \sup\{|g(t) - h(t)|: t \in [0, 1]|\} \\
-&= d(f, g) + d(g, h).
-\end{align}\end{split}\]
-Example 5:
-Suppose \((f_1, f_2, \dots), (g_1, g_2, \dots), (h_1, h_2, \dots) \in V,\) where \(V\) is the set of sequences \((a_1, a_2, \dots)\) of real numbers such that \(\sum_{n=1}^\infty |a_k| < \infty.\)
-Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((f_1, f_2, \dots), (h_1, h_2, \dots)) = \\
-&= \sum_{n=1}^\infty |f_k - h_k| \\
-&= \sum_{n=1}^\infty |(f_k - g_k) + (g_k - h_k)| \\
-&\leq \sum_{n=1}^\infty |f_k - g_k| + |g_k - h_k| \\
-&= d((f_1, f_2, \dots), (g_1, g_2, \dots)) + d((g_1, g_2, \dots), (h_1, h_2, \dots)).
-\end{align}\end{split}\]
-
-
-
-Exercise 6.A.2
-Prove that every finite subset of a metric space is closed.
-
-
-Solution
-
-
-
-
-Suppose \((V, d)\) is a metric space and let \(A\) be a finite subset of \(V.\)
-Let \(x \in A'.\)
-Since \(A\) is finite, the set \(\{d(x, y) \in \mathbb{R}^+: y \in A\}\) has a minimum element, and this minimum element is non-zero, say equal to some positive \(m \in \mathbb{R}\)
-Therefore, any open ball centered on \(x\) with radius less than or equal to \(m\) is contained in \(A'.\)
-Since \(x\) was arbitrary, \(A'\) is open, so \(A\) is closed.
-
-
-
-Exercise 6.A.3
-Prove that every closed ball in a metric space is closed.
-
-
-Solution
-
-
-
-
-Suppose \((V, d)\) is a metric space.
-Let \(r \in \mathbb{R}^+\) and \(f \in V.\)
-Let \(g \in \overline{B}(f, r)'.\)
-Then, \(d(f, g) > r\) so the open ball \(B(g, d(f, g) - r)\) does not intersect the closed ball \(\overline{B}(f, r).\)
-[Otherwise there would exist a common element \(h \in B(g, d(f, g) - r) \cap \overline{B}(f, r)\) which leads to a contradiction via the triangle inequality since \(d(f, g) \leq d(f, h) + d(h, g) < r.\)]
-Since the two balls do not intersect, \(B(g, r - d(f, g)) \subseteq \overline{B}(f, r)',\) which means that \(\overline{B}(f, r)'\) is an open set, so \(\overline{B}(f, r)\) is a closed set.
-
-
-
-Exercise 6.A.4
-Suppose \(V\) is a metric space.
-
-Prove that the union of each collection of open subsets of \(V\) is an open subset of \(V.\)
Prove that the intersection of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-
-
-
-
-In the following parts, \(V\) is a metric space and \(\mathcal{A}\) is a collection of subsets of \(V.\)
-Part 1:
-Let \(S = \cup_{A \in \mathcal{A}} A.\)
-If \(s \in S,\) there exists some \(A \in \mathcal{A}\) such that \(s \in A.\)
-Since every \(A \in \mathcal{A}\) is open, there exists an open ball centered on \(s\) that is contained in \(A.\)
-This open ball is also contained in \(S,\) so \(S\) is open.
-Part 2:
-Now suppose that, in addition, \(\mathcal{A} = \{A_1, \dots A_N\}\) is finite.
-If \(s \in S,\) then \(s \in A_n\) for \(n = 1, \dots, N.\)
-Since each \(A\) is open, for each \(A_n \in \mathcal{A},\) there exists an open ball centered on \(s\) with radius \(r_n,\) which is contained in \(A_n.\)
-Now, letting \(r = \min\{r_1, \dots, r_N\}\) we see that the open ball centered on \(s\) with radius \(r\) is contained in each \(A_n\) and therefore it is also contained in their intersection, i.e. it is contained in \(S.\)
-Therefore there exists an open ball centered on \(s \in S,\) which implies that \(S\) is open.
-
-
-
-Exercise 6.A.5
-Suppose \(V\) is a metric space.
-
-Prove that the intersection of each collection of closed subsets of \(V\) is an open subset of \(V.\)
Prove that the union of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-
-
-
-
-The complement of an intersection of a collection of sets is equal to the union of the complements of the sets in the collection.
-Similarly, the complement of a union of a collection of sets is equal to the intersection of the complements of the sets in the collection.
-Therefore, applying the result of the previous exercise we arrive at the two required results.
-
-
-
-Exercise 6.A.6
-
-Prove that if \(V\) is a metric space, \(f \in V,\) and \(r > 0,\) then \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
Give an example of a metric space \(V,\) \(f \in V,\) and \(r > 0\) such that \(\overline{B(f, r)} \neq \overline{B}(f, r).\)
-
-
-Solution
-
-
-
-
-Part 1:
-First, note that \(B(f, r) \subseteq \overline{B}(f, r).\)
-Second \(\overline{B}(f, r)\) is a closed set, and the closure of a set is the intersection of all closed sets that contain it, so \(\overline{B(f, r)}\) is contained in any closed set that contains \(B(f, r),\) so \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
-Part 2:
-Let \(V = \mathbb{Z}\) with the metric \(d: \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}^+\) defined as \(d(f, g) = |f - g|.\)
-Also let \(f = 0\) and \(r = 1.\)
-Then \(B(f, r) = \{0\}\) and so \(\overline{B(f, r)} = \{0\}.\)
-However, \(\overline{B}(f, r) = \{-1, 0, 1\}\) so \(\overline{B(f, r)} \neq \overline{B}(f, r)\) as required.
-
-
-
-Exercise 6.A.7
-Show that each sequence in a metric space has at most one limit.
-
-
-Solution
-
-
-
-
-Let \((V, d)\) be a metric space.
-Suppose \(f_1, f_2, \dots \in V\) is a sequence in \(V.\)
-If \(a, b \in V\) are limits of \(f_1, f_2, \dots,\) then
-
-\[\begin{equation}
-\lim_{n \to \infty} d(f_n, a) = 0 \text{ and } \lim_{n \to \infty} d(f_n, b) = 0.
-\end{equation}\]
-From the triangle inequality, \(d(a, b) \leq d(a, f_n) + d(f_n, b),\) and taking limits of both sides, we conclude that \(d(a, b) = 0,\) which implies that \(a = b.\)
-Therefore the sequence can have at most one limit in \(V.\)
-
-
-
-Exercise 6.A.8
-Prove that each open subset of a metric space \(V\) is the union of some sequence of closed subsets of \(V.\)
-
-
-Solution
-
-
-
-
-Let \(V\) be a metric space, let \(U\) be an open subset of \(V\) and define
-
-\[\begin{equation}
-U_r = \{f \in V: d(f, g) \geq r \text{ for all } g \in U'\}.
-\end{equation}\]
-Therefore \(U_r\) is the set of elements in \(U\) that are at least a distance \(r\) away from \(V.\)
-Note also that \(U_r \subseteq U.\)
-Now, note that for fixed \(g \in U',\) the set \(\{f \in V: d(f, g) \geq r\}\) is closed, because it is the complement of the open set \(\{f \in V: d(f, g) < r\}.\)
-The intersection of a collection of closed sets is closed, so \(U_r\) is closed.
-Now, note that for each \(n \in \mathbb{Z}^+,\) we have \(U_{\frac{1}{n}} \subseteq U,\) from which it follows that \(\bigcup_{n = 1}^\infty U_{\frac{1}{n}} \subseteq U.\)
-Conversely, if \(x \in U,\) since \(U\) is open there exists a ball of radius \(\frac{1}{n}\) for some \(n \in \mathbb{Z}^+\) contained in \(U,\) so \(x \in U_{\frac{1}{n}},\) from which it follows that \(U \subseteq \bigcup_{n = 1}^\infty U_{\frac{1}{n}}.\)
-Therefore \(U\) is the union of a sequence of closed sets in \(V,\) as required.
-
-
-
-Exercise 6.A.10
-Prove or give a counterexample:
-If \(V\) is a metric space and \(U, W\) are subserts of \(V,\) then \(\overline{U} \cup \overline{W} = \overline{U \cup W}.\)
-
-
-Solution
-
-
-
-
-If \(v \in \overline{U},\) then there exists a sequence of elements in \(U\) whose limit is \(v.\)
-Therefore there exists a sequence of elements in \(U \cup W\) whose limit is \(v,\) so \(v \in \overline{U \cup W}.\)
-Similarly, if \(v \in \overline{W},\) it follows that \(v \in \overline{U \cup W}.\)
-We conclude that \(\overline{U} \cup \overline{W} \subseteq \overline{U \cup W}.\)
-If \(v \in \overline{U \cup W},\) then \(v\) must be the limit of a sequence of elements in \(U \cup W.\)
-This sequence must have a infinite subsequence of elements in at least one of \(U\) or \(W\) with \(v\) as its limit, so \(v \in \overline{U} \cup \overline{W}.\)
-We conclude that \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W},\) which completes the proof.
-
-
-
-Exercise 6.A.11
-Prove or give a counterexample:
-If \(V\) is a metric space and \(U, W\) are subsets of \(V,\) then \(\overline{U} \cap \overline{W} = \overline{U \cap W}.\)
-
-
-Solution
-
-
-
-
-The equation does not hold.
-As a counterexample, consider \(\mathbb{R}\) with the metric
-
-\[d(f, g) = |f - g|,\]
-and let \(U = (-1, 0), W = (0, 1).\)
-Then we have \(\overline{U} = [-1, 0]\) and \(\overline{W} = [0, 1].\)
-Therefore, we have \(\overline{U} \cap \overline{W} = \{0\},\) but \(U \cap W = \emptyset\) so \(\overline{U \cap W} = \emptyset.\)
-
-
-
-Exercise 6.A.12
-Suppose \((U, d_U), (V, d_V)\) and \((W, d_W)\) are metric spaces.
-Suppose also that \(T: U \to V\) and \(S: V \to W\) are continuous functions.
-
-Using the definition of continuity, show that \(S \circ T: U \to W\) is continuous.
Using the equivalence of continuity with the property that the limit of a function is equal to the function of its limit, show that \(S \circ T: U \to W\) is continuous.
Using the equivalence of continuity with the property that the inverse image of an open set under a function is open, show that \(S \circ T: U \to W\) is continuous.
-
-
-Solution
-
-
-
-
-Part 1:
-Let \(v \in V\) and \(\epsilon > 0.\)
-Since \(S\) is continuous, there exists \(\delta_S > 0\) such that \(d_W(S(v), S(v')) < \epsilon\) for all \(v' \in V\) such that \(d_V(v, v') < \delta_S.\)
-Let \(u \in U.\)
-Since \(T\) is continuous, there exists \(\delta_T > 0\) such that \(d_V(T(u), T(u')) < \delta_S\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T.\)
-Letting \(v = T(u)\) and putting together these facts, we see that \(d_W(S \circ T(u), S \circ T(u')) < \epsilon\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T,\) which shows that \(S \circ T\) is continuous.
-Part 2:
-Suppose \(u_1, u_2, \dots\) is a sequence in \(U\) with limit \(u \in U.\)
-Since \(S\) and \(T\) are both continuous
-
-\[\begin{split}\begin{align}
-S \circ T(u) &= S(T(u)) \\
-&= S\left(T\left(\lim_{k \to \infty} u_k\right)\right) \\
-&= S\left(\lim_{k \to \infty} T(u_k)\right) \\
-&= \lim_{k \to \infty} S\left(T(u_k)\right) \\
-&= \lim_{k \to \infty} S \circ T (u_k)
-\end{align}\end{split}\]
-Therefore \(S \circ T\) is continuous.
-Part 3:
-Suppose \(G\) is an open subset in \(W.\)
-Since \(S\) is continuous, \(S^{-1}(G)\) is open in \(V\) and since \(T\) is continuous, \(T^{-1}(S^{-1}(G)) = (S \circ T)^{-1}(G)\) is open in \(U,\) so \(S \circ T\) is continuous.
-
-
-
-Exercise 6.A.14
-Suppose a Cauchy sequence in a metric space has a convergent subsequence.
-Prove that the Cauchy sequence converges.
-
-
-Solution
-
-
-
-
-Let \((V, d)\) be a metric space and \(v_1, v_2, \dots\) be a Cauchy sequence in \(V.\)
-Suppose that \(v_{k_1}, v_{k_2}, \dots\) is a subsequence which converges to \(v \in V.\)
-Let \(\epsilon > 0.\)
-Since the sequence \(v_1, v_2, \dots\) is Cauchy, there exists \(K\) such that for all \(k, k' \geq K\) we have \(d(v_k, v_{k'}) < \frac{\epsilon}{2}.\)
-In addition, by our earlier assumption, there exists \(N\) such that for all \(n \geq N\) we have \(d(v_{k_n}, v) < \frac{\epsilon}{2}.\)
-Then, letting \(L\) be the maximum of \(K\) and \(k_N\) we see that for any \(l \geq L\) it holds that \(d(v_l, v) \leq d(v_l, v_k) + d(v_k, v) < \epsilon,\) concluding the proof.
-
-
-
-Exercise 6.A.16
-Suppose \((U, d)\) is a metric space.
-Let \(W\) denote the set of all Cauchy sequences of elements of \(U.\)
-
-For \((f_1, f_2, \dots)\) and \((g_1, g_2, \dots)\) in \(W,\) define \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots)\) to mean that \(\lim_{k \to \infty} d(f_k, g_k) = 0.\)
-Show that \(\equiv\) is an equivalence relation on \(W.\)
Let \(V\) denote the set of equivalence classses of elements of \(W\) under the equivalence relation above.
-For \((f_1, f_2, \dots) \in W,\) let \((f_1, f_2, \dots)\hat{~}\) denote the equivalence class of \((f_1, f_2, \dots).\)
-Show that the following definition of \(d_V: V \times V \to [0, \infty)\) makes sense and that \(d_V\) is a metric on \(V\)
-
-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k).
-\end{equation}\]
-
-Show that \((V, d_V)\) is a complete metric space.
Show that the map from \(U\) to \(V\) that takes \(f \in U\) to \((f, f, f, \dots)\hat{~}\) preserves distances, meaning that for all \(f, g \in U,\) we have
-
-\[d(f, g) = d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~})\]
-
-Explain why (4) shows that every metric space is a subset of some complete metric space.
-
-
-Solution
-
-
-
-
-Part 1:
-First, we have that \(d(f_k, f_k) = 0\) so \((f_1, f_2, \dots) \equiv (f_1, f_2, \dots).\)
-Second, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) we have
-
-\[\lim_{k \to \infty} d(f_k, g_k) = 0 \implies \lim_{k \to \infty} d(g_k, f_k) = 0,\]
-so it follows that \((g_1, g_2, \dots) \equiv (f_1, f_2, \dots).\)
-Third, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) and \((g_1, g_2, \dots) \equiv (h_1, h_2, \dots),\) we have that
-
-\[\lim_{k \to \infty} d(f_k, h_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) = 0,\]
-which means that \((f_1, f_2, \dots) \equiv (h_1, h_2, \dots).\)
-Therefore \(\equiv\) is an equivalence relation.
-Part 2:
-First, for \(d_V\) to be well-defined, it should not matter which representative elements of \((f_1, f_2, \dots)\hat{~}\) and \((g_1, g_2, \dots)\hat{~}\) we pick in the right hand side of the equation that defines \(d_V.\)
-In particular, suppose \((\tilde{f}_1, \tilde{f}_2, \dots) \in (f_1, f_2, \dots)\hat{~}.\)
-Then
-
-\[\begin{equation}
-\lim_{k \to \infty} d(\tilde{f}_k, g_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(f_k, \tilde{f}_k)) = \lim_{k \to \infty} d(f_k, g_k),
-\end{equation}\]
-so it does not matter which representative element of \((f_1, f_2, \dots)\hat{~}\) we pick when defining \(d_V.\)
-The same holds for \((g_1, g_2, \dots)\hat{~},\) so \(d_V\) is well-defined.
-Second, we show that \(d_V\) is a metric.
-By the definition of \(d_V,\) we have that
-
-\[d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) \geq 0,\]
-with equality holding if and only if \(\lim_{k \to \infty} d(f_k, g_k) = 0,\) which in turn holds if and only if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) which is equivalent to \((f_1, f_2, \dots)\hat{~} = (g_1, g_2, \dots)\hat{~}.\)
-In addition, we have that
-
-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = d_V((g_1, g_2, \dots)\hat{~}, (f_1, f_2, \dots)\hat{~})
-\end{equation}\]
-because \(d(f_k, g_k) = d(g_k, f_k).\)
-Lastly, if \((h_1, h_2, \dots) \in W,\) we have that
-
-\[\begin{split}\begin{align}
-&~~~~d_V((f_1, f_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}) =\\
-&= \lim_{k \to \infty} d(f_k, h_k) \\
-&\leq\lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) \\
-&= d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) + d_V((g_1, g_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}),
-\end{align}\end{split}\]
-so \(d_V\) satisfies the triangle inequality, completing the proof that it is a metric.
-Part 3:
-Suppose that \((v_1, v_2, \dots)\) is a Cauchy sequence in \(V.\)
-We will show that there exists an element \(w \in V\) and \(K \in \mathbb{Z}^+\) such that for all \(k \geq K\) we have \(\lim_{i \to \infty} d((v_k)_i, w_i) = 0.\)
-Since each sequence \(v_k\) is itself Cauchy, there exists \(N_k \in \mathbb{Z}^+\) such that \(N_k \geq k\) and for all \(i, j \geq N_k,\) we have \(d(v_i, v_j) < \frac{1}{k}.\)
-Define the terms of the sequence \(w\) to be \(w_k = (v_k)_{N_k}.\)
-We now show that \(w\) is in fact the limit of \((v_1, v_2, \dots).\)
-Let \(\epsilon > 0.\)
-Since \((v_1, v_2, \dots)\) is Cauchy, there exists \(N_{\epsilon}^{(1)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{\epsilon}^{(1)},\) we have
-
-\[\lim_{k \to \infty} d((v_i)_k, (v_j)_k) < \frac{\epsilon}{3}.\]
-In addition, for fixed \(k \in \mathbb{Z}^+\) since each sequence \(v_k\) is Cauchy, there exists \(N_{k, \epsilon}^{(2)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{k, \epsilon}^{(2)},\) we have
-
-\[\lim_{k \to \infty} d((v_k)_i, (v_k)_j) < \frac{\epsilon}{3}.\]
-Now, we have that
-
-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, w_i) \\
-&= d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_{N_i}) \\
-&\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_j) + d((v_i)_j, (v_i)_{N_i})
-\end{align}\end{split}\]
-where in the first line we have applied the triangle inequality, in the second line we have substituted the definition \(w_i = (v_i)_{N_i}\) and in the third line we have again used the triangle inequality again.
-Now, letting \(i, k \geq N_{\epsilon}^{(1)}\) means that the limit of the second term in the inequality, as \(j \to \infty\), is smaller than \(\epsilon / 3.\)
-In addition, letting \(i, j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)})\) means that the first term in the inequality is smaller than \(\epsilon / 3.\)
-Lastly, by the definition of \(N_i,\) letting \(j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)}, N_i)\) means that the last term in the inequality is smaller than \(\frac{1}{i}.\)
-We therefore obtain
-
-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq \lim_{j \to \infty} d((v_k)_i, (v_k)_j) + \lim_{j \to \infty} d((v_k)_j, (v_i)_j) + \lim_{j \to \infty} d((v_i)_j, (v_i)_{N_i}) \\
-&< \frac{2\epsilon}{3} + \frac{1}{i}
-\end{align}\end{split}\]
-so for all \(i > \frac{3}{\epsilon}\) we have \(d((v_k)_i, w_i) < \epsilon.\)
-This means that \(d_V(v_k\hat{~}, w\hat{~}) < \epsilon\) for all \(k \geq N_{\epsilon}^{(1)}\) which means that \(w\hat{~}\) is the limit of the sequence \((v_1\hat{~}, v_2\hat{~}, \dots),\) so \(V\) is a complete metric space.
-Part 4:
-By part (2), defining \(f_k = f\) and \(g_k = g\) for all \(k \in \mathbb{Z}^+,\) we have that
-
-\[d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k) = d(f, g),\]
-as required.
-Part 5:
-We can add elements to the set \(U\) to ensure that the resulting set is complete.
-Specifically, we add to \(U\) the set
-\(S = \left\{w \in W | w \neq (f, f, f, \dots)\hat{~} \text{ for any } f \in U \right\},\)
-to obtain the larger set \(\hat{U} = U \cup S.\)
-Now, for \(u \in \hat{U},\) we define \(\hat{u}\) to be equal to \((u, u, \dots)\hat{~}\) if \(u \in U\) and equal to \(u\) otherwise.
-Then, define the function \(d_\hat{U}: \hat{U} \times \hat{U} \to \mathbb{R}\) as
-
-\[d_\hat{U}(u, v) = d_V(\hat{u}, \hat{v}).\]
-This function is a metric over \(\hat{U}.\)
-Further, if \((u_1, u_2, \dots)\) is a Cauchy sequence of elements of \(U,\) then \((\hat{u_1}, \hat{u_2}, \dots)\) is a Cauchy sequence of equivalence classes in \(V.\)
-Since \(V\) is complete, the sequence \((\hat{u_1}, \hat{u_2}, \dots)\) has a limit, denoted \(\hat{u},\) in \(V.\)
-Therefore the sequence \((u_1, u_2, \dots)\) converges to \(\hat{u}\) in the metric space \((\hat{U}, d_{hat{U}}).\)
-
-
-
Exercise 6.A.1
-Verify that each of the following examples of sets \(V\) and functions \(d: V \times V \to \mathbb{R}\) are indeed metric spaces.
--
-
Suppose \(V\) is a nonempty set. -Define \(d\) as
-
-\[\begin{split}\begin{equation}
-d(f, g) = \begin{cases}
-1 &\text{ if } f = g \\
-0 &\text{ if } f \neq g
-\end{cases}
-\end{equation}\end{split}\]
--
-
Let \(V = \mathbb{R}.\) -Define \(d\) as
-
-\[\begin{equation}
-d(x, y) = |x - y|.
-\end{equation}\]
--
-
Let \(V = \mathbb{R}.\) -For \(n \in \mathbb{Z}^+,\) define \(d\) as
-
-\[\begin{equation}
-d((x_1, \dots, x_n), (y_1, \dots, y_n)) = \max\{\|x_1 - y_1|, \dots, |x_n - y_n|\}.
-\end{equation}\]
--
-
Let \(V = C([0, 1]),\) the set of continuous real-valued functions on \([0, 1].\) -Define \(d\) as
-
-\[\begin{equation}
-d(f, g) = \sup\{|f(t) - g(t)|: t \in [0, 1]|\}.
-\end{equation}\]
--
-
Let \(V\) be the set of sequences \((a_1, a_2, \dots)\) with \(a_k \in \mathbb{R}\) and \(\sum_{n=1}^\infty |a_k| < \infty.\) -Define \(d\) as
-
-\[\begin{equation}
-d((a_1, a_2, \dots), (b_1, b_2, \dots)) = \sum_{n=1}^\infty |a_k - b_k|.
-\end{equation}\]
-
-
-Solution
-Solution
-
-
-
-
-
-For all the definitions above, \(d(f, g) \geq 0\) and equality with zero holds only if \(f, g.\) -Further, \(d(f, g) = d(g, f).\) -It remains to show the triangle inequality
-
-\[\begin{equation}
-d(f, h) \leq d(f, g) + d(g, h) \text{ for all } f, g, h \in V
-\end{equation}\]
-holds for each example.
-Example 1: -Suppose \(f, g, h \in \mathbb{V}.\) -If \(f = h,\) then the triangle inequality holds since
-
-\[\begin{equation}
-d(f, h) = 0 \leq d(f, g) + d(g, h),
-\end{equation}\]
-holds. -If \(f \neq h,\) then at least one of \(f \neq g\) and \(g \neq h\) must hold, so the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = 1 \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 2: -Suppose \(f, g, h \in \mathbb{R}.\) -Then the triangle inequality holds because
-
-\[\begin{equation}
-d(f, h) = |f - h| = |(f - g) + (g - h)| \leq |f - g| + |g - h| \leq d(f, g) + d(g, h).
-\end{equation}\]
-Example 3: -Suppose \(n \in \mathbb{Z}^+\) and \((x_1, \dots, x_n), (y_1, \dots, y_n), (z_1, \dots, z_n) \in \mathbb{R}^n.\) -Then the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((x_1, \dots, x_n), (z_1, \dots, z_n)) = \\
-&= \max\{|x_1 - z_1|, \dots, |x_n - z_n|\} \\
-&= \max\{|(x_1 - y_1) + (y_1 - z_1)|, \dots, |(x_n - y_n) + (y_n - z_n)|\} \\
-&\leq \max\{|x_1 - y_1| + |y_1 - z_1|, \dots, |x_n - y_n| + |y_n - z_n|\} \\
-&\leq \max\{|x_1 - y_1|, \dots, |x_n - y_n|\} + \max\{|y_1 - z_1|, \dots, |y_n - z_n|\} \\
-&\leq d((x_1, \dots, x_n), (y_1, \dots, y_n)) + d((y_1, \dots, y_n), (z_1, \dots, z_n)).
-\end{align}\end{split}\]
-Example 4: -Suppose \(f, g, h \in C([0, 1]).\) -Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-d(f, h) &= \sup\{|f(t) - h(t)|: t \in [0, 1]|\} \\
-&= \sup\{|(f(t) - g(t)) + (g(t) - h(t))|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)| + |g(t) - h(t)|: t \in [0, 1]|\} \\
-&\leq \sup\{|f(t) - g(t)|: t \in [0, 1]|\} + \sup\{|g(t) - h(t)|: t \in [0, 1]|\} \\
-&= d(f, g) + d(g, h).
-\end{align}\end{split}\]
-Example 5: -Suppose \((f_1, f_2, \dots), (g_1, g_2, \dots), (h_1, h_2, \dots) \in V,\) where \(V\) is the set of sequences \((a_1, a_2, \dots)\) of real numbers such that \(\sum_{n=1}^\infty |a_k| < \infty.\) -Then, the triangle inequality holds because
-
-\[\begin{split}\begin{align}
-&~~~~d((f_1, f_2, \dots), (h_1, h_2, \dots)) = \\
-&= \sum_{n=1}^\infty |f_k - h_k| \\
-&= \sum_{n=1}^\infty |(f_k - g_k) + (g_k - h_k)| \\
-&\leq \sum_{n=1}^\infty |f_k - g_k| + |g_k - h_k| \\
-&= d((f_1, f_2, \dots), (g_1, g_2, \dots)) + d((g_1, g_2, \dots), (h_1, h_2, \dots)).
-\end{align}\end{split}\]
-Exercise 6.A.2
-Prove that every finite subset of a metric space is closed.
-
-
-Solution
-Solution
-
-
-
-
-
-Suppose \((V, d)\) is a metric space and let \(A\) be a finite subset of \(V.\) -Let \(x \in A'.\) -Since \(A\) is finite, the set \(\{d(x, y) \in \mathbb{R}^+: y \in A\}\) has a minimum element, and this minimum element is non-zero, say equal to some positive \(m \in \mathbb{R}\) -Therefore, any open ball centered on \(x\) with radius less than or equal to \(m\) is contained in \(A'.\) -Since \(x\) was arbitrary, \(A'\) is open, so \(A\) is closed.
-Exercise 6.A.3
-Prove that every closed ball in a metric space is closed.
-
-
-Solution
-Solution
-
-
-
-
-
-Suppose \((V, d)\) is a metric space. -Let \(r \in \mathbb{R}^+\) and \(f \in V.\) -Let \(g \in \overline{B}(f, r)'.\) -Then, \(d(f, g) > r\) so the open ball \(B(g, d(f, g) - r)\) does not intersect the closed ball \(\overline{B}(f, r).\) -[Otherwise there would exist a common element \(h \in B(g, d(f, g) - r) \cap \overline{B}(f, r)\) which leads to a contradiction via the triangle inequality since \(d(f, g) \leq d(f, h) + d(h, g) < r.\)] -Since the two balls do not intersect, \(B(g, r - d(f, g)) \subseteq \overline{B}(f, r)',\) which means that \(\overline{B}(f, r)'\) is an open set, so \(\overline{B}(f, r)\) is a closed set.
-Exercise 6.A.4
-Suppose \(V\) is a metric space.
--
-
Prove that the union of each collection of open subsets of \(V\) is an open subset of \(V.\)
-Prove that the intersection of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-Solution
-
-
-
-
-
-In the following parts, \(V\) is a metric space and \(\mathcal{A}\) is a collection of subsets of \(V.\)
-Part 1: -Let \(S = \cup_{A \in \mathcal{A}} A.\) -If \(s \in S,\) there exists some \(A \in \mathcal{A}\) such that \(s \in A.\) -Since every \(A \in \mathcal{A}\) is open, there exists an open ball centered on \(s\) that is contained in \(A.\) -This open ball is also contained in \(S,\) so \(S\) is open.
-Part 2: -Now suppose that, in addition, \(\mathcal{A} = \{A_1, \dots A_N\}\) is finite. -If \(s \in S,\) then \(s \in A_n\) for \(n = 1, \dots, N.\) -Since each \(A\) is open, for each \(A_n \in \mathcal{A},\) there exists an open ball centered on \(s\) with radius \(r_n,\) which is contained in \(A_n.\) -Now, letting \(r = \min\{r_1, \dots, r_N\}\) we see that the open ball centered on \(s\) with radius \(r\) is contained in each \(A_n\) and therefore it is also contained in their intersection, i.e. it is contained in \(S.\) -Therefore there exists an open ball centered on \(s \in S,\) which implies that \(S\) is open.
-Exercise 6.A.5
-Suppose \(V\) is a metric space.
--
-
Prove that the intersection of each collection of closed subsets of \(V\) is an open subset of \(V.\)
-Prove that the union of each finite collection of open subsets of \(V\) is an open subset of \(V.\)
-
-
-Solution
-Solution
-
-
-
-
-
-The complement of an intersection of a collection of sets is equal to the union of the complements of the sets in the collection. -Similarly, the complement of a union of a collection of sets is equal to the intersection of the complements of the sets in the collection. -Therefore, applying the result of the previous exercise we arrive at the two required results.
-Exercise 6.A.6
--
-
Prove that if \(V\) is a metric space, \(f \in V,\) and \(r > 0,\) then \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
-Give an example of a metric space \(V,\) \(f \in V,\) and \(r > 0\) such that \(\overline{B(f, r)} \neq \overline{B}(f, r).\)
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-Solution
-Solution
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-Part 1: -First, note that \(B(f, r) \subseteq \overline{B}(f, r).\) -Second \(\overline{B}(f, r)\) is a closed set, and the closure of a set is the intersection of all closed sets that contain it, so \(\overline{B(f, r)}\) is contained in any closed set that contains \(B(f, r),\) so \(\overline{B(f, r)} \subseteq \overline{B}(f, r).\)
-Part 2: -Let \(V = \mathbb{Z}\) with the metric \(d: \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}^+\) defined as \(d(f, g) = |f - g|.\) -Also let \(f = 0\) and \(r = 1.\) -Then \(B(f, r) = \{0\}\) and so \(\overline{B(f, r)} = \{0\}.\) -However, \(\overline{B}(f, r) = \{-1, 0, 1\}\) so \(\overline{B(f, r)} \neq \overline{B}(f, r)\) as required.
-Exercise 6.A.7
-Show that each sequence in a metric space has at most one limit.
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-Solution
-Solution
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-Let \((V, d)\) be a metric space. -Suppose \(f_1, f_2, \dots \in V\) is a sequence in \(V.\) -If \(a, b \in V\) are limits of \(f_1, f_2, \dots,\) then
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-\[\begin{equation}
-\lim_{n \to \infty} d(f_n, a) = 0 \text{ and } \lim_{n \to \infty} d(f_n, b) = 0.
-\end{equation}\]
-From the triangle inequality, \(d(a, b) \leq d(a, f_n) + d(f_n, b),\) and taking limits of both sides, we conclude that \(d(a, b) = 0,\) which implies that \(a = b.\) -Therefore the sequence can have at most one limit in \(V.\)
-Exercise 6.A.8
-Prove that each open subset of a metric space \(V\) is the union of some sequence of closed subsets of \(V.\)
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-Solution
-Solution
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-Let \(V\) be a metric space, let \(U\) be an open subset of \(V\) and define
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-\[\begin{equation}
-U_r = \{f \in V: d(f, g) \geq r \text{ for all } g \in U'\}.
-\end{equation}\]
-Therefore \(U_r\) is the set of elements in \(U\) that are at least a distance \(r\) away from \(V.\) -Note also that \(U_r \subseteq U.\) -Now, note that for fixed \(g \in U',\) the set \(\{f \in V: d(f, g) \geq r\}\) is closed, because it is the complement of the open set \(\{f \in V: d(f, g) < r\}.\) -The intersection of a collection of closed sets is closed, so \(U_r\) is closed.
-Now, note that for each \(n \in \mathbb{Z}^+,\) we have \(U_{\frac{1}{n}} \subseteq U,\) from which it follows that \(\bigcup_{n = 1}^\infty U_{\frac{1}{n}} \subseteq U.\) -Conversely, if \(x \in U,\) since \(U\) is open there exists a ball of radius \(\frac{1}{n}\) for some \(n \in \mathbb{Z}^+\) contained in \(U,\) so \(x \in U_{\frac{1}{n}},\) from which it follows that \(U \subseteq \bigcup_{n = 1}^\infty U_{\frac{1}{n}}.\) -Therefore \(U\) is the union of a sequence of closed sets in \(V,\) as required.
-Exercise 6.A.10
-Prove or give a counterexample: -If \(V\) is a metric space and \(U, W\) are subserts of \(V,\) then \(\overline{U} \cup \overline{W} = \overline{U \cup W}.\)
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-Solution
-Solution
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-If \(v \in \overline{U},\) then there exists a sequence of elements in \(U\) whose limit is \(v.\) -Therefore there exists a sequence of elements in \(U \cup W\) whose limit is \(v,\) so \(v \in \overline{U \cup W}.\) -Similarly, if \(v \in \overline{W},\) it follows that \(v \in \overline{U \cup W}.\) -We conclude that \(\overline{U} \cup \overline{W} \subseteq \overline{U \cup W}.\)
-If \(v \in \overline{U \cup W},\) then \(v\) must be the limit of a sequence of elements in \(U \cup W.\) -This sequence must have a infinite subsequence of elements in at least one of \(U\) or \(W\) with \(v\) as its limit, so \(v \in \overline{U} \cup \overline{W}.\) -We conclude that \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W},\) which completes the proof.
-Exercise 6.A.11
-Prove or give a counterexample: -If \(V\) is a metric space and \(U, W\) are subsets of \(V,\) then \(\overline{U} \cap \overline{W} = \overline{U \cap W}.\)
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-Solution
-Solution
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-The equation does not hold. -As a counterexample, consider \(\mathbb{R}\) with the metric
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-\[d(f, g) = |f - g|,\]
-and let \(U = (-1, 0), W = (0, 1).\) -Then we have \(\overline{U} = [-1, 0]\) and \(\overline{W} = [0, 1].\) -Therefore, we have \(\overline{U} \cap \overline{W} = \{0\},\) but \(U \cap W = \emptyset\) so \(\overline{U \cap W} = \emptyset.\)
-Exercise 6.A.12
-Suppose \((U, d_U), (V, d_V)\) and \((W, d_W)\) are metric spaces. -Suppose also that \(T: U \to V\) and \(S: V \to W\) are continuous functions.
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Using the definition of continuity, show that \(S \circ T: U \to W\) is continuous.
-Using the equivalence of continuity with the property that the limit of a function is equal to the function of its limit, show that \(S \circ T: U \to W\) is continuous.
-Using the equivalence of continuity with the property that the inverse image of an open set under a function is open, show that \(S \circ T: U \to W\) is continuous.
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-Solution
-Solution
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-Part 1: -Let \(v \in V\) and \(\epsilon > 0.\) -Since \(S\) is continuous, there exists \(\delta_S > 0\) such that \(d_W(S(v), S(v')) < \epsilon\) for all \(v' \in V\) such that \(d_V(v, v') < \delta_S.\) -Let \(u \in U.\) -Since \(T\) is continuous, there exists \(\delta_T > 0\) such that \(d_V(T(u), T(u')) < \delta_S\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T.\) -Letting \(v = T(u)\) and putting together these facts, we see that \(d_W(S \circ T(u), S \circ T(u')) < \epsilon\) for all \(u' \in U\) such that \(d_U(u, u') < \delta_T,\) which shows that \(S \circ T\) is continuous.
-Part 2: -Suppose \(u_1, u_2, \dots\) is a sequence in \(U\) with limit \(u \in U.\) -Since \(S\) and \(T\) are both continuous
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-\[\begin{split}\begin{align}
-S \circ T(u) &= S(T(u)) \\
-&= S\left(T\left(\lim_{k \to \infty} u_k\right)\right) \\
-&= S\left(\lim_{k \to \infty} T(u_k)\right) \\
-&= \lim_{k \to \infty} S\left(T(u_k)\right) \\
-&= \lim_{k \to \infty} S \circ T (u_k)
-\end{align}\end{split}\]
-Therefore \(S \circ T\) is continuous.
-Part 3: -Suppose \(G\) is an open subset in \(W.\) -Since \(S\) is continuous, \(S^{-1}(G)\) is open in \(V\) and since \(T\) is continuous, \(T^{-1}(S^{-1}(G)) = (S \circ T)^{-1}(G)\) is open in \(U,\) so \(S \circ T\) is continuous.
-Exercise 6.A.14
-Suppose a Cauchy sequence in a metric space has a convergent subsequence. -Prove that the Cauchy sequence converges.
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-Solution
-Solution
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-Let \((V, d)\) be a metric space and \(v_1, v_2, \dots\) be a Cauchy sequence in \(V.\) -Suppose that \(v_{k_1}, v_{k_2}, \dots\) is a subsequence which converges to \(v \in V.\) -Let \(\epsilon > 0.\) -Since the sequence \(v_1, v_2, \dots\) is Cauchy, there exists \(K\) such that for all \(k, k' \geq K\) we have \(d(v_k, v_{k'}) < \frac{\epsilon}{2}.\) -In addition, by our earlier assumption, there exists \(N\) such that for all \(n \geq N\) we have \(d(v_{k_n}, v) < \frac{\epsilon}{2}.\) -Then, letting \(L\) be the maximum of \(K\) and \(k_N\) we see that for any \(l \geq L\) it holds that \(d(v_l, v) \leq d(v_l, v_k) + d(v_k, v) < \epsilon,\) concluding the proof.
-Exercise 6.A.16
-Suppose \((U, d)\) is a metric space. -Let \(W\) denote the set of all Cauchy sequences of elements of \(U.\)
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For \((f_1, f_2, \dots)\) and \((g_1, g_2, \dots)\) in \(W,\) define \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots)\) to mean that \(\lim_{k \to \infty} d(f_k, g_k) = 0.\) -Show that \(\equiv\) is an equivalence relation on \(W.\)
-Let \(V\) denote the set of equivalence classses of elements of \(W\) under the equivalence relation above. -For \((f_1, f_2, \dots) \in W,\) let \((f_1, f_2, \dots)\hat{~}\) denote the equivalence class of \((f_1, f_2, \dots).\) -Show that the following definition of \(d_V: V \times V \to [0, \infty)\) makes sense and that \(d_V\) is a metric on \(V\)
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-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k).
-\end{equation}\]
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Show that \((V, d_V)\) is a complete metric space.
-Show that the map from \(U\) to \(V\) that takes \(f \in U\) to \((f, f, f, \dots)\hat{~}\) preserves distances, meaning that for all \(f, g \in U,\) we have
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-\[d(f, g) = d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~})\]
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Explain why (4) shows that every metric space is a subset of some complete metric space.
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-Solution
-Solution
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-Part 1: -First, we have that \(d(f_k, f_k) = 0\) so \((f_1, f_2, \dots) \equiv (f_1, f_2, \dots).\) -Second, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) we have
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-\[\lim_{k \to \infty} d(f_k, g_k) = 0 \implies \lim_{k \to \infty} d(g_k, f_k) = 0,\]
-so it follows that \((g_1, g_2, \dots) \equiv (f_1, f_2, \dots).\) -Third, if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) and \((g_1, g_2, \dots) \equiv (h_1, h_2, \dots),\) we have that
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-\[\lim_{k \to \infty} d(f_k, h_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) = 0,\]
-which means that \((f_1, f_2, \dots) \equiv (h_1, h_2, \dots).\) -Therefore \(\equiv\) is an equivalence relation.
-Part 2: -First, for \(d_V\) to be well-defined, it should not matter which representative elements of \((f_1, f_2, \dots)\hat{~}\) and \((g_1, g_2, \dots)\hat{~}\) we pick in the right hand side of the equation that defines \(d_V.\) -In particular, suppose \((\tilde{f}_1, \tilde{f}_2, \dots) \in (f_1, f_2, \dots)\hat{~}.\) -Then
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-\[\begin{equation}
-\lim_{k \to \infty} d(\tilde{f}_k, g_k) \leq \lim_{k \to \infty} (d(f_k, g_k) + d(f_k, \tilde{f}_k)) = \lim_{k \to \infty} d(f_k, g_k),
-\end{equation}\]
-so it does not matter which representative element of \((f_1, f_2, \dots)\hat{~}\) we pick when defining \(d_V.\) -The same holds for \((g_1, g_2, \dots)\hat{~},\) so \(d_V\) is well-defined.
-Second, we show that \(d_V\) is a metric. -By the definition of \(d_V,\) we have that
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-\[d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) \geq 0,\]
-with equality holding if and only if \(\lim_{k \to \infty} d(f_k, g_k) = 0,\) which in turn holds if and only if \((f_1, f_2, \dots) \equiv (g_1, g_2, \dots),\) which is equivalent to \((f_1, f_2, \dots)\hat{~} = (g_1, g_2, \dots)\hat{~}.\) -In addition, we have that
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-\[\begin{equation}
-d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) = d_V((g_1, g_2, \dots)\hat{~}, (f_1, f_2, \dots)\hat{~})
-\end{equation}\]
-because \(d(f_k, g_k) = d(g_k, f_k).\) -Lastly, if \((h_1, h_2, \dots) \in W,\) we have that
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-\[\begin{split}\begin{align}
-&~~~~d_V((f_1, f_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}) =\\
-&= \lim_{k \to \infty} d(f_k, h_k) \\
-&\leq\lim_{k \to \infty} (d(f_k, g_k) + d(g_k, h_k)) \\
-&= d_V((f_1, f_2, \dots)\hat{~}, (g_1, g_2, \dots)\hat{~}) + d_V((g_1, g_2, \dots)\hat{~}, (h_1, h_2, \dots)\hat{~}),
-\end{align}\end{split}\]
-so \(d_V\) satisfies the triangle inequality, completing the proof that it is a metric.
-Part 3: -Suppose that \((v_1, v_2, \dots)\) is a Cauchy sequence in \(V.\) -We will show that there exists an element \(w \in V\) and \(K \in \mathbb{Z}^+\) such that for all \(k \geq K\) we have \(\lim_{i \to \infty} d((v_k)_i, w_i) = 0.\) -Since each sequence \(v_k\) is itself Cauchy, there exists \(N_k \in \mathbb{Z}^+\) such that \(N_k \geq k\) and for all \(i, j \geq N_k,\) we have \(d(v_i, v_j) < \frac{1}{k}.\) -Define the terms of the sequence \(w\) to be \(w_k = (v_k)_{N_k}.\) -We now show that \(w\) is in fact the limit of \((v_1, v_2, \dots).\)
-Let \(\epsilon > 0.\) -Since \((v_1, v_2, \dots)\) is Cauchy, there exists \(N_{\epsilon}^{(1)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{\epsilon}^{(1)},\) we have
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-\[\lim_{k \to \infty} d((v_i)_k, (v_j)_k) < \frac{\epsilon}{3}.\]
-In addition, for fixed \(k \in \mathbb{Z}^+\) since each sequence \(v_k\) is Cauchy, there exists \(N_{k, \epsilon}^{(2)} \in \mathbb{Z}^+\) such that for all \(i, j \geq N_{k, \epsilon}^{(2)},\) we have
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-\[\lim_{k \to \infty} d((v_k)_i, (v_k)_j) < \frac{\epsilon}{3}.\]
-Now, we have that
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-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, w_i) \\
-&= d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_{N_i}) \\
-&\leq d((v_k)_i, (v_k)_j) + d((v_k)_j, (v_i)_j) + d((v_i)_j, (v_i)_{N_i})
-\end{align}\end{split}\]
-where in the first line we have applied the triangle inequality, in the second line we have substituted the definition \(w_i = (v_i)_{N_i}\) and in the third line we have again used the triangle inequality again. -Now, letting \(i, k \geq N_{\epsilon}^{(1)}\) means that the limit of the second term in the inequality, as \(j \to \infty\), is smaller than \(\epsilon / 3.\) -In addition, letting \(i, j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)})\) means that the first term in the inequality is smaller than \(\epsilon / 3.\) -Lastly, by the definition of \(N_i,\) letting \(j \geq \max(N_{\epsilon}^{(1)}, N_{k, \epsilon}^{(2)}, N_i)\) means that the last term in the inequality is smaller than \(\frac{1}{i}.\) -We therefore obtain
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-\[\begin{split}\begin{align}
-d((v_k)_i, w_i) &\leq \lim_{j \to \infty} d((v_k)_i, (v_k)_j) + \lim_{j \to \infty} d((v_k)_j, (v_i)_j) + \lim_{j \to \infty} d((v_i)_j, (v_i)_{N_i}) \\
-&< \frac{2\epsilon}{3} + \frac{1}{i}
-\end{align}\end{split}\]
-so for all \(i > \frac{3}{\epsilon}\) we have \(d((v_k)_i, w_i) < \epsilon.\) -This means that \(d_V(v_k\hat{~}, w\hat{~}) < \epsilon\) for all \(k \geq N_{\epsilon}^{(1)}\) which means that \(w\hat{~}\) is the limit of the sequence \((v_1\hat{~}, v_2\hat{~}, \dots),\) so \(V\) is a complete metric space.
-Part 4: -By part (2), defining \(f_k = f\) and \(g_k = g\) for all \(k \in \mathbb{Z}^+,\) we have that
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-\[d_V((f, f, f, \dots)\hat{~}, (g, g, g, \dots)\hat{~}) = \lim_{k \to \infty} d(f_k, g_k) = d(f, g),\]
-as required.
-Part 5: -We can add elements to the set \(U\) to ensure that the resulting set is complete. -Specifically, we add to \(U\) the set
-\(S = \left\{w \in W | w \neq (f, f, f, \dots)\hat{~} \text{ for any } f \in U \right\},\)
-to obtain the larger set \(\hat{U} = U \cup S.\) -Now, for \(u \in \hat{U},\) we define \(\hat{u}\) to be equal to \((u, u, \dots)\hat{~}\) if \(u \in U\) and equal to \(u\) otherwise. -Then, define the function \(d_\hat{U}: \hat{U} \times \hat{U} \to \mathbb{R}\) as
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-\[d_\hat{U}(u, v) = d_V(\hat{u}, \hat{v}).\]
-This function is a metric over \(\hat{U}.\) -Further, if \((u_1, u_2, \dots)\) is a Cauchy sequence of elements of \(U,\) then \((\hat{u_1}, \hat{u_2}, \dots)\) is a Cauchy sequence of equivalence classes in \(V.\) -Since \(V\) is complete, the sequence \((\hat{u_1}, \hat{u_2}, \dots)\) has a limit, denoted \(\hat{u},\) in \(V.\) -Therefore the sequence \((u_1, u_2, \dots)\) converges to \(\hat{u}\) in the metric space \((\hat{U}, d_{hat{U}}).\)
-Epoch 10: loss 0.089 (train 0.024), acc. 0.973 (train 0.972)
+Epoch 5: loss 0.081 (train 0.045), acc. 0.959 (train 0.956)
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Conclusion#
-We have looked at the details of the transformer architecture. -It consists of identical blocks, each of which contains a self-attention and a multi-layer perceptron operation, together with normalisation layers and residual connections. -Coupling these together with position embeddings and an appropriate tokenisation layer makes up the entire transformer architecture. -We looked at a specific example for computer vision, the ViT, and trained it on MNIST.
+We have looked at the details of the Swin transformer. +The Swin transformer amounts to modifying the attention operation in a regular transformer by adding windows. +All tokens within a window attend to one another and there is no attention across windows. +To allow for information to propagate across windows, Swin applies window shifting between consecutive transformer blocks. +We looked at a specific example for computer vision and trained it on MNIST.
References#
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+[LLC+21]
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+Ze Liu, Yutong Lin, Yue Cao, Han Hu, Yixuan Wei, Zheng Zhang, Stephen Lin, and Baining Guo. Swin transformer: hierarchical vision transformer using shifted windows. In Proceedings of the IEEE/CVF international conference on computer vision, 10012–10022. 2021.
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