Problem Set 2
Table of contents
Problem 1
Let $ x_n \to x $ and $ y_n \to y $ be two convergent sequences in $ (X, d)$ . Then, $ d(x_n, y_n) \to d(x, y) $ .
Let $ {x_n}$ and $ {y_n}$ be Cauchy sequences in $ X$ . Show that $ d(x_n, y_n)$ converges.
Hint: show that in a metric space, $ \vert d(a, b) + d(a’, b’)\vert \leq d(a, a’) + d(b, b’)$ .
Remark 1. You may not assume $ x_n$ and $ y_n$ converges in the second part of this problem. This is only true in a Cauchy complete space.
(1.) Let $ \epsilon>0$ , $ \exists N_1 > 0$ , for $ n \geq N_1$ , $ d(x_n, x) < \epsilon / 2$ . $ \exists N_2 >0$ , for $ n \geq N_2$ , $ d(y_n, y) < \epsilon / 2.$
If $ n \geq \max{N_1, N_2}$ , then
\[d(x_n, y_n) \leq d(x_n, x) + d(x, y) + d(y, y_n)\leq d(x, y) + \epsilon\] \[d(x, y) \leq d(x, x_n) + d(x_n, y_n) + d(y_n, y)\leq d(x_n, y_n) + \epsilon\]Therefore,
\[\vert d(x_n, y_n) - d(x, y)\vert < \epsilon \Longrightarrow d(x_n, y_n) \to d(x, y)\](2.) Claim: $ \vert d(a, b) - d(a’, b’)\vert < d(a, a’) + d(b, b’).$
Proof: Since $ d(a, b) \leq d(a, a’) + d(a’, b’) + d(b, b),$ then $ d(a, b) - d(a’, b’) \leq d(a, a’) + d(b, b).$ Also $ d(a’, b’) \leq d(a’, a) + d(a, b) + d(b, b’),$ then $ d(a’, b) - d(a, b) \leq d(a, a’) + d(b, b’).$ Thus, $ \vert d(a, b) - d(a’, b’)\vert \leq d(a, a’) + d(b, b’).$
Let $ \epsilon>0$ . $ \exists M_1 > 0$ , for $ n > M_1$ and $ m > M_1$ , then $ d(x_n, x_m) < \epsilon / 2$ . $ \exists M_2 > 0$ , for all $ n > M_2$ , $ m > M_2$ , then $ d(y_n, y_m) < \epsilon / 2$ .
Denote $ M := \max{M_1, M_2}$ , for all $ n > M, m > M$ ,
\[\vert d(x_n, y_n) - d(x_m, y_m)\vert \leq d(x_n, y_n) + d(x_m, y_m) = \epsilon.\]Thus, \(\{d(x_n, y_n)\}_{n=1}^\infty\) is a Cauchy sequence in $\mathbb{R}$ . Since $ \mathbb{R}$ is Cauchy complete, \(\{d(x_n, y_n)\}_{n=1}^\infty\) converges.
Problem 2
In class, we have defined a set $ A \subset X $ to be closed if its complement is an open set in $ X $ . There is another useful definition of a closed set however. Show that $ A \subset X $ is closed if and only if every convergent sequence in $ A $ converges in $ A $ . In other words, if $ {x_n} $ is a convergent sequence in $ A $ such that $ x_n \to x $ , then $ x \in A $ .
We show the two definitions are equivalent.
(i) $ A $ ‘s complement is an open set in $ X $ .
(ii) If $ {x_n} $ is a convergent sequence in $ A $ such that $ x_n \to x $ , then $ x \in A $ .
(i) $ \implies$ (ii). $ \exists \varepsilon > 0, \, \text{s.t.}\, B(x,\varepsilon) = { y \in X : d(x,y) < \varepsilon } \subset A$ . If $ {x_n}$ is a convergent sequence, let $ x := \lim_{n \to \infty} x_n $ . Given $ \varepsilon > 0$ , $ \exists N \in \mathbb{N}$ , $ \forall n > N$ , $ d(x_n, x) < \varepsilon $ .
Since $ x_n \in A $ , then $ B(x_n, \varepsilon) \subset A $ . Thus $ x \in B(x_n, \varepsilon) \subset A $ .
(ii) $ \implies$ (i). Prove by contrapositive. Assume $ \forall \varepsilon > 0$ , there $ \exists y \in B(x,\varepsilon)$ , such that $ y \notin A $ , given $ x \in A $ . Let $ {x_n}$ be a convergent sequence in $ A $ , such that $ x_n \to x $ . Then $ \exists N \in \mathbb{N}$ , for $ n > N $ , $ d(x_n, x) < \varepsilon /2 $ . Since $ y \in B(x, \varepsilon) $ , $ d(x,y) < \varepsilon /2 $ . Thus,
\[d(x_n, y) \leq d(x_n, x) + d(x, y) < \varepsilon/2 + \varepsilon/2 = \varepsilon\]So $ x_n \to y $ . However $ y \notin A $ , which contradicts with the condition (ii). Therefore (i) holds.
Problem 3
Here, we will show that $ C^0([0,1]) $ is Cauchy complete with respect to the uniform distance. Suppose that $ f_n \in C^0([0,1]) $ is a Cauchy sequence. The uniform distance on $ C^0([a,b]) $ is defined as
\[d(f,g) = \max_{x \in [a,b]} \vert f(x) - g(x)\vert .\](a) Fix an arbitrary $ x_0 \in [0,1] $ . Show that $ \lim_{n \to \infty} f_n(x_0) $ exists.
Hint: $ \mathbb{R} $ is Cauchy complete.
Fix any $ x_0 \in [0,1] $ . Then for $ m,n \in \mathbb{N} $ ,
\(\vert f_n(x_0) - f_m(x_0)\vert \le \max_{x\in[0,1]} \vert f_n(x) - f_m(x)\vert .\)
Thus $ (f_n(x_0)) $ is a Cauchy sequence in $ \mathbb{R} $ . Since $ \mathbb{R} $ is complete, the limit $ \lim_{n\to\infty} f_n(x_0)$ exists for each $ x_0 \in [0,1] $ .
(b) Define $ f : [0,1] \to \mathbb{R} $ by
\(f(x) = \lim_{n \to \infty} f_n(x).\)
Show that for all $ \varepsilon > 0 $ , there exists $ N \in \mathbb{N} $ such that
\(\vert f_n(x) - f(x)\vert \le \varepsilon\)
for all $ x \in [0,1] $ and for all $ n \ge N $ .
Define a function $ f:[0,1]\to \mathbb{R} $ by
\[f(x) := \lim_{n\to\infty} f_n(x) \quad \text{for all } x \in [0,1].\]We claim that for every $ \varepsilon>0 $ there exists $ N\in\mathbb{N} $ such that for all $ n \ge N $ ,
\[\vert f_n(x) - f(x)\vert \le \varepsilon \quad \text{for all } x \in [0,1].\]Indeed, since $ {f_n} $ is Cauchy in the uniform distance, there exists $ N $ such that for all $ m,n \ge N $ ,
\[\max_{x\in[0,1]} \vert f_n(x) - f_m(x)\vert < \varepsilon.\]Fixing $ n \ge N $ and letting $ m \to \infty $ gives
\[\vert f_n(x) - f(x)\vert \le \varepsilon \quad \text{for all } x \in [0,1].\](c) Show that $ f(x) $ is continuous on $ [0,1]$ . I.e., $ f \in C^0([0,1]) $ .
Hint: To show $ f(x) $ is continuous at $ x_0 $ , consider
\(\vert f(x) - f(x_0)\vert \le \vert f(x) - f_n(x)\vert + \vert f_n(x) - f_n(x_0)\vert + \vert f_n(x_0) - f(x_0)\vert .\)
We now show $ f $ is continuous on $ [0,1]$ . Fix $ x_0 \in [0,1] $ and $ \varepsilon>0 $ . From part (b), there exists $ N_1 $ such that for all $ n \ge N_1 $ , for all $ x\in[0,1]$ , $ \vert f_n(x) - f(x)\vert < \varepsilon/3$ .
There exists $ N_2 $ such that for all $ n \ge N_2 $ , $ \vert f_n(x_0) - f(x_0)\vert < \varepsilon/3$
Pick $ n \ge \max{N_1,N_2} $ . Since $ f_n\in C^0([0,1]) $ , there exists $ \delta > 0 $ such that for all $ x $ with $ \vert x-x_0\vert <\delta $ , $ \vert f_n(x) - f_n(x_0)\vert < \varepsilon/3$ .
Then for such $ x $ ,
\(\begin{aligned} \vert f(x) - f(x_0)\vert &\le \vert f(x) - f_n(x)\vert + \vert f_n(x) - f_n(x_0)\vert + \vert f_n(x_0) - f(x_0)\vert \\ &< \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon. \end{aligned}\)
Thus $ f $ is continuous at $ x_0 $ . Since $ x_0 $ was arbitrary, $ f \in C^0([0,1]) $ .
(d) Using parts (a)–(c), explain why $ \lim_{n \to \infty} f_n = f $ as a sequence in $ C^0([0,1]) $ .
Finally, we show $ f_n \to f $ in the uniform distance. Let $ \varepsilon>0 $ . Since $ (f_n) $ is Cauchy in the uniform distance, there exists $ N $ such that for all $ m,n \ge N $ ,
\(\max_{x\in[0,1]} \vert f_n(x) - f_m(x)\vert < \varepsilon.\)
Fix $ n \ge N $ . Then for all $ x \in [0,1] $ ,
\(\vert f_n(x) - f(x)\vert = \lim_{m\to\infty} \vert f_n(x) - f_m(x)\vert \le \varepsilon.\)
Thus $ d(f_n(x),f(x))=\max_{x\in[0,1]} \vert f_n(x) - f(x)\vert \le \varepsilon$ . Hence $ f_n \to f $ .
Problem 4
Let $ \Vert \cdot\Vert $ be a norm on a vector space $ V $ , and let $ d(x,y) = \Vert x-y\Vert $ for all $ x, y \in V $ . Show the following three properties:
(a) $ d(\lambda x, \lambda y) = \vert \lambda\vert d(x,y) $ for all $ \lambda \in \mathbb{R} $ , and for all $ x,y \in V $ .
(b) Translation invariance: $ d(x+z, y+z) = d(x,y) $ for all $ x,y,z \in V $ .
(c) Prove $ d $ is a metric on $ V $ . This metric is called the metric induced by the norm.
(a) \(d(\lambda x, \lambda y) = \Vert \lambda x - \lambda y\Vert = \Vert \lambda (x-y)\Vert = \vert \lambda\vert \Vert x-y\Vert = \vert \lambda\vert d(x,y)\)
(b)
\(d(x+z, y+z) = \Vert (x+z) - (y+z)\Vert = \Vert x-y\Vert = d(x,y)\)(c) (i) Positive definite: $ d(x,y) \ge 0.$ If $ x=y $ , then $ x-y=0 $ , so $ d(x,y)=0 $ . If $ d(x,y)=0 $ , then $ \Vert x-y\Vert =0 $ , hence $ x=y $ .
(ii) Symmetry: $ d(x,y) = \Vert x-y\Vert = \Vert y-x\Vert = d(y,x)$
(iii) Triangle inequality: \(d(x,z) = \Vert x-z\Vert \le \Vert x-y\Vert + \Vert y-z\Vert = d(x,y) + d(y,z)\)
Problem 5
Let $ U $ be an open set in the metric space $ (X,d) $ . Show that $ U $ can be written as a union of arbitrarily many open balls.
Let $ U \subseteq X $ be open. $ \forall x \in U$ , $ \exists \varepsilon_x > 0$ such that the open ball
\(B(x,\varepsilon_x) := \{ y \in X : d(x,y) < \varepsilon_x \}\)satisfies $ B(x,\varepsilon_x) \subseteq U $ .
Claim: \(U = \bigcup_{x \in U} B(x,\varepsilon_x)\) Pf: $ (\subseteq)$ Let $ z \in U $ . By openness, there is some $ \varepsilon_z > 0$ such that $ B(z,\varepsilon_z) \subseteq U $ . In particular, $ z \in B(z,\varepsilon_z) \subseteq \bigcup_{x \in U} B(x,\varepsilon_x) $ .
$ (\supseteq)$ Conversely, if $ z \in B(x,\varepsilon_x) $ for some $ x \in U $ , then by construction $ B(x,\varepsilon_x) \subseteq U $ . Hence, $ z \in U $ .