KindiakMath

Category: Exercises

Arzelà-Ascoli Applied

Problem 1. Let $F : [0, 1]^2 \to \mathbb R$ be continuous. For each $x \in [0, 1]$ , define $f_x := F(x, \cdot) : [0, 1] \to \mathbb R$ . Prove that the set

$\mathcal F := \{ f_x \in \mathcal C([0, 1], \mathbb R) : x \in [0, 1] \}$

is compact.

(Click for Solution)

Solution. For boundedness, we note that for any $f_x$ ,

$\displaystyle \| f_x \|_\infty = \sup_{y \in K} | F(x,y) | \leq \sup_{(x,y) \in K \times K} |F(x,y)| = \|F\|_\infty,$

as required.

For closedness, fix any sequence $\{f_{x_n}\} \subseteq \mathcal F$ and suppose $f_{x_n} \to f$ . Since $\{x_n\} \subseteq [0, 1]$ is bounded, by the Bolzano-Weierstrass theorem, it contains a convergent subsequence $x_{n_k} \to x$ . Hence, for any $y \in [0, 1]$ , since $F(\cdot, y)$ is continuous,

$\begin{aligned} f(y) = \lim_{n \to \infty} f_{x_n}(y) &= \lim_{n \to \infty} F(x_{n}, y) \\ &= \lim_{k \to \infty} F(x_{n_k}, y) = F(x,y) = f_x(y), \end{aligned}$

so that $f = f_x$ , as required.

For equicontinuity, fix $\epsilon > 0$ . We will consider $K \times K$ equipped with the supremum metric (thus inducing the box topology) for computational simplicity. Fix $k > 0$ to be tuned. For each $(x, y) \in K \times K$ , there exists $\delta_{x,y} > 0$ such that

$\| (x,y) - (u,v) \|_\infty < \delta_{x,y} \quad \Rightarrow \quad |F(x,y) - F(u,v)| < k \cdot \epsilon.$

Let $d$ denote the supremum metric. Since $\{B_{d}((x,y), \delta_{x,y}/2) : (x,y) \in K \times K\}$ forms an open cover for the compact space $K \times K$ , we can extract a finite sub cover $\{B_d((x_i,y_i), \delta_i) : i = 1,\dots, n\}$ , where we denoted $\delta_i = \delta_{x_i,y_i}$ for brevity.

Choose $\delta = \min\{\delta_1,\dots,\delta_n\}/2$ . Now fix $r, s \in [0, 1]$ and suppose $|r - s| < \delta$ .

By the triangle inequality, there exists $i$ such that

$\max\{ \|(x, r) - (x, y_i)\|_\infty , \|(x, s) - (x, y_i)\|_\infty \} < \delta_i.$

Hence,

$\max \{ |F(x,r) - F(x,y_i)|, |F(x,s) - F(x,y_i)| \} < k \cdot \epsilon.$

Setting $k = 1/2$ and applying the triangle inequality yields

$|f_x(r) - f_x(s)| = |F(x,r) - F(x,s)| < \epsilon,$

as required.

By the Arzelà-Ascoli theorem, $\mathcal F$ is compact. Hence, every sequence $\{f_n\}$ in $\mathcal F$ has a convergent subsequence.

—Joel Kindiak, 30 May 25, 1247H

November 10, 2025
Tietze Extension Theorem

Let $K$ be a normal Hausdorff space and $C \subseteq K$ be closed. Assume that the Urysohn lemma is a valid means to construct useful continuous maps.

Problem 1. For any $r > 0$ and any continuous map $f : C \to [-r, r]$ , construct a continuous map $g : K \to [-r/3, r/3]$ such that $(g-f)(C) \in [0, 2r/3]$ .

(Click for Solution)

Solution. Let $f : C \to [-r, r]$ be a continuous map. We observe that $K \backslash C$ is open. We will first construct a function $g$ that approximates $f$ reasonably well.

Decompose $[-r, r] = [-r,-r/3] \cup [-r/3,r/3] \cup [r/3, r]$ and define the closed subspaces

$C_1 := f^{-1}([-r,-r/3]), \quad C_2 := f^{-1}([-r/3,r/3]), \quad C_3 := f^{-1}([r/3,r])$

of $K$ . Since $C_1 \cap C_3 = \emptyset$ and $K$ is normal, the Urysohn lemma gives us a continuous function $g : K \to [-r/3,r/3]$ such that $g(C_1) = \{-r/3\}$ and $g(C_3) = \{r/3\}$ .

We observe the following estimates: by construction, $|g(x)| \leq r/3$ for any $x \in K$ . By considering cases, $|g(x) - f(x)| \leq 2r/3$ for any $x \in K$ .

Problem 2. For any continuous map $f : C \to [-1, 1]$ , construct one extension $F : K \to [-1, 1]$ of $f$ such that $F$ is continuous and $F|_C = f$ .

(Click for Solution)

Solution. Let $f : C \to [-1, 1]$ be any continuous map. Use Problem 1 to construct $g_1 : K \to [-1/3,1/3]$ such that

$|(f-g_1)(x)| \leq 2/3,\quad x \in C.$

Replacing $f$ with $f-g_1$ , use Problem 1 again to construct $g_2 : K \to [-1/3^2, 1/3^2]$ such that

$|(f - (g_1 + g_2))(x)| \leq (2/3)^2,\quad x \in C.$

Repeating the process, we obtain a sequence $\{g_n\}$ of continuous maps such that $g_n : K \to [-1/3^n, 1/3^n]$ and

$|(f - (g_1 + g_2 + \dots + g_n))(x)| \leq (2/3)^n,\quad x \in C.$

Then $f_n := g_1 + \cdots + g_n$ should converge to some limit function $F$ , and by our estimates, since $d_\infty( f_n, f) \leq (2/3)^n$ when restricted to $C$ , we have $f_n \to f$ uniformly on $C$ too. Therefore, $F|_C = f$ , as required.

Problem 3. For any continuous map $f : C \to \mathbb R$ , construct one extension $F : K \to \mathbb R$ of $f$ such that $F$ is continuous and $F|_C = f$ .

(Click for Solution)

Solution. Since $(-1, 1) \cong \mathbb R$ via the homeomorphism $t \mapsto t/(1-t^2)$ , it suffices to prove the result for $f : C \to (-1, 1)$ . By Problem 2, extend $f : C \to (-1, 1) \subseteq [-1, 1]$ to a continuous map $g : K \to [-1, 1]$ .

Define the closed set $D := g^{-1}(\{-1, 1\})$ . Since $g$ extends $f$ , $g(C) \subseteq (-1, 1)$ Hence, $C \cap D = \emptyset$ . Use Urysohn’s lemma to supply a continuous map $\phi : K \to [0, 1]$ where $\phi(D) = \{0\}$ and $\phi(C) = \{1\}$ .

Now verify that the definition $F =\phi \cdot g : K \to (-1, 1)$ is the desired extension, i.e. $\pm 1 \notin F(K)$ and $F|_C = f$ .

—Joel Kindiak, 17 May 25, 1538H

September 19, 2025
Baire Category Theorem

In this exercise, we prove the Baire category theorem (i.e. Problem 3).

Let $K$ be a topological space. Recall that a subset $S \subseteq K$ is dense in $K$ if $\bar S = K$ .

Problem 1. Prove that $S$ is dense in $K$ if and only if for any nonempty open set $U \subseteq K$ , $S \cap U \neq \emptyset$ .

(Click for Solution)

Solution. We prove this statement by contrapositive. For the direction $(\Rightarrow)$ , suppose $\bar S \subsetneq K$ . Then there exists $x \in K \backslash \bar S$ . Since the latter is open, it contains a neighbourhood $U$ of $x$ so that $U \subseteq K \backslash \bar S \subseteq K \backslash S$ , yielding $S \cap U = \emptyset$ , as required.

For the direction $(\Leftarrow)$ , suppose there exists a nonempty open set $U \subseteq K$ such that $S \cap U = \emptyset$ . Then the closed set $K \backslash U$ contains $S$ and thus $\bar S \subseteq K \backslash U$ . Since $U \neq \emptyset$ , there exists $x \in U$ so that $x \notin K \backslash U \supseteq \bar S$ . Hence, $\bar S \subsetneq K \backslash U \subsetneq K$ , as required.

Now suppose $K$ is a complete metric space with metric $d$ .

Problem 2. Prove that if $C \subseteq K$ is closed, then $C$ is complete.

(Click for Solution)

Solution. Fix any Cauchy sequence $\{x_n\}$ in $C \subseteq K$ . Since $K$ is complete, $x_n \to x$ for some $x \in K$ . Since $C$ is closed and $K$ is a metric space, $x \in C$ . Since every Cauchy sequence in $C$ converges in $C$ , we have that $C$ is complete.

Problem 3. Prove that for any countable collection $\{U_n \}$ of open dense sets in $K$ , $\bigcap_n U_n$ is dense in $K$ .

(Click for Solution)

Solution. Fix any open set $V \subseteq K$ . Since each $U_k$ is dense, $V \cap U_k \neq \emptyset$ . It suffices to prove that $V \cap \bigcap_n U_n \neq \emptyset$ .

Define $U_0 := V$ . Inductively, since $B(x_k, r_k) \cap U_{k+1} \neq \emptyset$ , find $x_{k+1} \in B(x_k, r_k) \cap U_{k+1}$ and $r_{k+1} \in (0, 2^{-(k+1)})$ such that

$\overline{ B(x_{k+1}, r_{k+1}) } \subseteq B(x_k, r_k) \cap U_{k+1}.$

Inductively, for any $k$ ,

$\displaystyle \overline {B(x_{k+1}, r_{k+1})} \subseteq V \cap \bigcap_{i=1}^k U_i.$

Now consider the sequence $\{x_n\}$ . It is Cauchy since for sufficiently large $m, n$ , $d(x_m, x_n) < 2^{-\min\{m, n\}}$ . Since $\bar V$ is closed, it is complete. Hence, $\{x_n\}$ is Cauchy in $\bar V$ and converges to some $x \in \bar V$ . By a straightforward $\epsilon$ – $K$ verification, $x \in V \cap \bigcap_{n} U_n$ , as required.

Problem 4. Prove that if $K = \bigcup_{i=1}^\infty C_i$ for a countable collection $\{C_i\}$ of closed sets, then at least one $C_i$ has nonempty interior.

(Click for Solution)

Solution. By assumption, taking complements yields $\bigcap_{i=1}^\infty K \backslash C_i = \emptyset$ . For a contradiction, suppose all $C_i$ have empty interior. We claim that $\overline{ K \backslash C_i } = K$ , so that $K \backslash C_i$ being dense, via Problem 3, yields the desired contradiction $K = \overline{\bigcap_{i=1}^\infty K \backslash C_i} = \bar{\emptyset} = \emptyset$ .

Suppose for a contradiction that $\overline{ K \backslash C_i } \subsetneq K$ . Then there exists $x \in K \backslash \overline{ K \backslash C_i }$ . Since the latter is open, there exists a neighbourhood $U \subseteq K \backslash \overline{ K \backslash C_i }$ of $x$ . Hence, $K \backslash C_i \subseteq \overline{ K \backslash C_i } \subseteq K \backslash U$ implies that $C_i \supseteq U \ni x$ , contradicting the assumption that $C_i$ has empty interior.

—Joel Kindiak, 3 May 25, 1237H

August 29, 2025
Several Metrization Lemmas

Problem 0. Prove that the function $f : [0, \infty) \to \mathbb R$ defined by $f(x) = x/(1+x)$ is subadditive i.e. for any $r, s \in [0, \infty)$ , $f(r+s) \leq f(r) + f(s)$ with equality if and only if $rs = 0$ .

(Click for Solution)

Solution. Since $r , s \geq 0$ ,

$\begin{aligned} f(r+s) &= \frac{r+s}{1+(r+s)} \\ &= \frac r{1+(r+s)} + \frac s{1+(r+s)} \\ &\leq \frac r{1+r} + \frac s{1+s} = f(r) + f(s). \end{aligned}$

The equality condition holds since $f'(x) = 1/(1+x)^2 > 0$ so that $f$ is strictly increasing.

Problem 1. Let $K$ be metrizable with a metric $d$ . Prove that there exists a metric $\rho$ bounded by $1$ on $K$ that generates the same metric topology on $K$ .

(Click for Solution)

Solution. Define $\rho := f \circ d$ . Since $f([0, 1)) \subseteq [0, 1)$ , $f \circ d$ is bounded by $1$ . To check that the topologies generated are the same, let $\mathcal B_d$ and $\mathcal B_\rho$ denote the metric topologies induced by the metrics $d, \rho$ respectively.

To prove that $\mathcal T_d \subseteq \mathcal T_\rho$ , fix $B_d(x, \epsilon) \in \mathcal B_d$ . We can assume $\epsilon < 1$ without loss of generality. Since $f$ is injective, $B_\rho(x, f(\epsilon)) \subseteq B_d(x, \epsilon)$ , establishing $(\subseteq )$ . To prove that $\mathcal T_d \supseteq \mathcal T_\rho$ , fix $B_\rho(x, \epsilon) \in \mathcal B_\rho$ , assuming $\epsilon < 1$ once again. Then $B_d(x, f^{-1}(\epsilon)) \subseteq B_\rho(x, \epsilon)$ , establishes $(\supseteq )$ .

Problem 2. For each $n$ , let $K_n$ be metrizable with a bounded metric $\rho_n$ as made possible by Problem 1. Define $K := \prod_{n=1}^\infty K_n$ equipped with the product topology. Prove that the map $d : K \times K \to \mathbb R$ defined by

$\displaystyle d(\mathbf x, \mathbf y) := \sum_{n=1}^\infty \rho_n(x_n, y_n) \cdot 2^{-n}$

is a well-defined metric on $K$ and generates the product topology on $K$ .

(Click for Solution)

Solution. Since $\rho_n$ is bounded by $1$ , for any $\mathbf x, \mathbf y \in K$ , $d(\mathbf x, \mathbf y)$ converges absolutely and is bounded by $1$ too. Hence, $d$ is well-defined.

To check the product topology claim, fix any $\mathbf x \in K$ and basis element $B$ of $K$ that contains $\mathbf x$ . Assume $B = B_1(x_1,\epsilon_1) \times \cdots \times B_n(x_n,\epsilon_n) \times \prod_{i=n+1}^\infty K_i$ and $\epsilon_i < 1$ without loss of generality. Define $\epsilon = \min\{\epsilon_1 \cdot 2^{-1},\dots,\epsilon_n \cdot 2^{-n}\}$ , so that $B_d(\mathbf x, \epsilon) \subseteq B$ .

On the other hand, fix $\mathbf x \in K$ and $B_d(\mathbf x, \epsilon)$ , and assume $\epsilon < 1$ without loss of generality. Find $N$ such that $1/2^N < \epsilon/2$ . Define

$\displaystyle B = B_1(x_1,\epsilon/2) \times \cdots \times B_N(x_n,\epsilon/2) \times \prod_{i=N+1}^\infty K_i.$

Then $B \subseteq B_d(\mathbf x, \epsilon)$ since for any $\mathbf y \in B$ ,

$\displaystyle \begin{aligned} d(\mathbf x, \mathbf y) &:= \sum_{n=1}^\infty \rho_n(x_n, y_n) \cdot 2^{-n} \\ &\leq \frac{\epsilon}2 \cdot \sum_{n=1}^N 2^{-n} + \sum_{n=N+1}^\infty 2^{-n} \\ &< \frac{\epsilon}2 \cdot \frac {2^{-1}}{1 - 2^{-1}} + \frac{2^{-(N+1)}}{1 - 2^{-1}} \\ &= \frac {\epsilon}2 \cdot 1 + 2^{-N} < \frac{\epsilon}2 + \frac{\epsilon}2 = \epsilon.\end{aligned}$

Problem 3. Briefly explain why $\mathbb R^\omega$ is metrizable.

(Click for Solution)

Solution. Since each $\mathbb R$ is metrizable, by Problem 2, $\mathbb R^\omega := \prod_{n=1}^\infty \mathbb R$ equipped with the product topology is metrizable.

Problem 4. Let $K, L$ be topological spaces and $f : K \to L$ be a continuous injection that is a homeomorphism to $f(K)$ . Prove that if $L$ is metrizable, then so is $K$ .

(Click for Solution)

Solution. Given the metric $d$ on $L$ , define the metric $\rho$ on $K$ by $\rho(x, y) := d(f(x),f(y))$ and verify that $d$ generates the relevant topology.

Problem 5. Conclude that for any topological space $K$ , if a map $f : K \to \mathbb R^\omega$ described in Problem 4 exists, then $K$ is metrizable.

(Click for Solution)

Solution. By Problem 3, $\mathbb R^\omega$ is metrizable. By Problem 4, $K$ is metrizable.

—Joel Kindiak, 25 Apr 25, 1325H

August 8, 2025
Hausdorff Bookkeeping

In this exercise, we are going to verify various properties of Hausdorff spaces.

Problem 1. Suppose $K$ is a Hausdorff space. Prove that any subspace $L \subseteq K$ equipped with the subspace topology is also Hausdorff.

(Click for Solution)

Solution. Recall that $U \subseteq L$ is open if and only if there exists an open set $V \subseteq K$ such that $U = L \cap V$ . Let $x_1,x_2 \in L$ be distinct points. Since $K$ is Hausdorff, there exist disjoint neighbourhoods $V_i$ of $x_i$ in $K$ . Then $U_i := L \cap V_i$ are disjoint neighbourhoods of $x_i$ in $L$ , as required.

Problem 2. Let $I$ be any index set and $\{K_\alpha\}_{\alpha \in I}$ be a collection of Hausdorff spaces. Prove that $\prod_{\alpha \in I} K_\alpha$ is Hausdorff in the product topology.

(Click for Solution)

Solution. Fix distinct points $\mathbf x, \mathbf y \in \prod_{\alpha \in I} K_\alpha$ . Then there exists $\alpha \in I$ such that $\mathbf x_\alpha \neq \mathbf y_\alpha$ . Find disjoint neighbourhoods $U_\alpha, V_\alpha$ of $\mathbf x_\alpha, \mathbf y_\alpha$ respectively. Define the disjoint neighbourhoods $U = \prod_{\beta} W_\beta, V = \prod_{\beta} X_\beta$ of $\mathbf x, \mathbf y$ by

$W_\beta := \begin{cases} U_\beta & \alpha = \beta, \\ K_\beta & \alpha \neq \beta, \end{cases}\quad X_\beta := \begin{cases} V_\beta & \alpha = \beta, \\ K_\beta & \alpha \neq \beta. \end{cases}$

as required.

Problem 3. Prove that $K$ is a Hausdorff space if and only if its diagonal

$\Delta_K := \{t \times t : t \in K\} \subseteq K \times K =: K^2$

is closed.

(Click for Solution)

Solution. For the direction $(\Rightarrow)$ , suppose $K$ is a Hausdorff space. Fix $s \times t \in K^2 \backslash \Delta_K$ . Since $s \times t \notin \Delta_K$ , $s \neq t$ . Hence, there exist disjoint neighbourhoods $U, V$ of $s, t$ respectively. Then $U \times V$ is a neighbourhood of $s \times t$ .

We claim that $U \times V \subseteq K^2 \backslash \Delta_K$ . Suppose otherwise. Then there exists $s' \times t' \in (U \times V) \cap \Delta_K$ . Hence, $s' = t'$ . Thus, $\emptyset = U \cap V = \{s'\}$ , a contradiction. Therefore, $U \times V \subseteq K^2 \backslash \Delta_K$ . Since $s \times t$ has been chosen arbitrarily, $K^2 \backslash \Delta_K$ is open, so that $\Delta_K$ is closed, as required.

For the direction $(\Leftarrow)$ , suppose $\Delta_K$ is closed. Then $K^2 \backslash \Delta_K$ is open. Fix distinct $s, t \in K$ . Then $s \times t \in K^2 \backslash \Delta_K$ . Thus, there exist open sets $U,V \subseteq K$ such that $s \times t \in U \times V \subseteq K^2 \backslash \Delta_K$ . We observe that if $U \cap V$ is not empty, then $U \times V \not\subseteq K^2 \backslash \Delta_K$ , a contradiction. Therefore, $U \cap V = \emptyset$ , as required.

—Joel Kindiak, 30 Mar 2025, 2340H

May 23, 2025
The Metric Topology

Recall that for any metric space $(K, d)$ , we can define open balls with centre $x$ and radius $r > 0$ by

$B_d(x,r) := \{u \in K : d(x, u) < r\}.$

We have previously seen that the collection $\mathcal B := \{ B_d(x,r) : x \in K, r > 0\}$ forms a topological basis for $K$ and induces the metric topology $\mathcal T_d$ on $K$ .

Problem 1. For any $L \subseteq K$ , define the metric subspace $(L, d_L)$ by $d_L = d|_{L \times L}$ . Prove that $\mathcal T_{d_L}$ coincides with the subspace topology of $(K, \mathcal T_d)$ .

(Click for Solution)

Solution. Regarding $L \subseteq K$ as a topological subspace, the subspace topology is generated by a basis of the form

$\mathcal B_1 := \{B_d(x,r) \cap L : x \in K, r > 0\}.$

On the other hand, the metric topology induced by $d_L$ is generated by a basis of the form

$\mathcal B_2 := \{B_{d_L}(x, r) : x \in L, r > 0\}.$

We aim to prove that $\mathcal B_1$ generates $\mathcal T_{d_L}$ and $\mathcal B_2$ generates the subspace topology.

For the first claim, fix $B_{d_L}(x, r) \in \mathcal B_2$ and $y \in B_{d_L}(x, r)$ . We seek $r_1 > 0$ such that

$y \in B_d(y, r_1) \cap L \subseteq B_{d_L}(x, r).$

Observe that for any $w \in B_d(y, r_1) \cap L$ ,

$d_L(x,w) \leq d_L(x,y) + d_L(y,w) = d_L(x,y) + d(y,w) < d_L(x,y) + r_1.$

Settin $r_1 := r - d_L(x,y) > 0$ yields the desired result. Hence, $\mathcal B_1$ generates $\mathcal T_{d_L}$ .

For the second claim, fix any nonempty $B_d(x,r) \cap L \in \mathcal B_1$ and $y \in B_d(x,r) \cap L$ , that is, $y \in L$ and $d(x,y) < r$ . We seek $r_2 > 0$ such that $B_{d_L}(y, r_2) \subseteq B_d(x,r)$ . Now, for any $u \in B_{D_L}(y, r_2)$ ,

$d(u,x) \leq d(u,y) + d(y, x) = d_L(u, y) + d(y, x) < r_2 + d(y, x).$

Setting $r_2 := r - d(y,x) > 0$ therefore yields the desired result. Hence, $\mathcal B_2$ generates the subspace topology.

Problem 2. Prove that any inner product space $(V, \langle \cdot , \cdot \rangle)$ on $\mathbb R$ is also a metric space, with a metric induced by $\langle \cdot , \cdot \rangle$ . Deduce that the map $d : \mathbb R^n \times \mathbb R^n \to \mathbb R^n$ defined by

$\displaystyle d(\mathbf u, \mathbf v) := \left( \sum_{k=1}^n |u_k - v_k|^2 \right)^{1/2}$

is a metric on $\mathbb R^n$ , called the Euclidean metric on $\mathbb R^n$ .

(Click for Solution)

Solution. The inner product induces a norm $\| \cdot \| : V \to \mathbb R$ on $V$ via the map $\| \mathbf v \| := \sqrt{\langle \mathbf v, \mathbf v \rangle}$ . The norm, in turn, induces a metric $d : V \times V \to \mathbb R$ on $V$ via the map $d(\mathbf u, \mathbf v) := \|\mathbf u - \mathbf v\|$ . For the final claim, we use the definition of the inner product on $\mathbb R^n$ to deduce that

$\begin{aligned} d(\mathbf u, \mathbf v) &= \| \mathbf u - \mathbf v \| \\ &= \sqrt{\langle \mathbf u - \mathbf v, \mathbf u - \mathbf v \rangle } \\ &= \left( \sum_{k=1}^n (u_k - v_k)^2 \right)^{1/2}.\end{aligned}$

Problem 3. A function $f : \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}$ is subadditive if

$f(x+y) \leq f(x) + f(y), \quad x , y \in \mathbb R_{\geq 0}.$

Prove that $\sqrt{\cdot} : \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}$ is subadditive.

(Click for Solution)

Solution. For nonnegative $a, b$ , since $\sqrt{\cdot}$ is bijective and non-decreasing,

$a+b = (\sqrt{a})^2 + (\sqrt{b})^2 \leq (\sqrt a+ \sqrt b)^2.$

Taking square roots, $\sqrt{a+b} \leq \sqrt a + \sqrt b$ .

Problem 4. For metric spaces $(K_1, d_1)$ and $(K_2, d_2)$ and $K := K_1 \times K_2$ , prove that the map $d : K \times K \to \mathbb R$ defined by

$d((u_1, u_2), (v_1, v_2)) := \sqrt{ d_1(u_1, v_1)^2 + d_2(u_2,v_2)^2 }$

is a metric on $K_1 \times K_2$ .

(Click for Solution)

Solution. For the the nontrivial triangle inequality property, denote

$\begin{aligned} A &:= d((u_1, u_2), (v_1, v_2)) \\ B &:= d((u_1, u_2), (w_1, w_2)) \\ C &:= d((w_1, w_2), (v_1, v_2)). \end{aligned}$

We assert that $A \leq B + C$ . By bookkeeping and using the triangle inequality for each $d_i$ ,

$\begin{aligned} A^2 \leq B^2 + C^2 + 2(d_1(u_1,w_1)d_1(w_1,v_1) + d_2(u_2,w_2)d_2(w_2,v_2)).\end{aligned}$

By the Cauchy-Schwarz inequality,

$\displaystyle d_1(u_1,w_1)d_1(w_1,v_1) + d_2(u_2,w_2)d_2(w_2,v_2) \leq \sqrt{B^2 C^2} = BC.$

Hence,

$A^2 \leq B^2 + C^2 + 2BC = (B+C)^2.$

Taking square roots yields the desired result.

Problem 5. For metric spaces $(K_1, d_1)$ and $(K_2, d_2)$ and the metric $d$ on $K_1 \times K_2$ as defined in Problem 4, prove that $\mathcal T_d$ coincides with the product topology on $K_1 \times K_2$ .

(Click for Solution)

Solution. Recall that the product topology is generated by a basis given by

$\mathcal B_1 := \{B_{d_1}(x_1,r_1) \times B_{d_2}(x_2,r_2) : x_i \in K_i, r_i > 0\}$

On the other hand, the topology generated by $d$ has a basis of the form

$\mathcal B_2 := \{B_d(\mathbf x, r) : \mathbf x \in K_1 \times K_2, r > 0\}.$

We claim that $\mathcal T_{\mathcal B_1} = \mathcal T_{B_2}$ . To prove $(\subseteq)$ , fix $B_{d_1}(x_1,r_1) \times B_{d_2}(x_2,r_2) \in \mathcal B_1$ and any element $(y_1,y_2)$ it contains. We seek $r > 0$ such that

$B_d((y_1,y_2), r) \subseteq B_{d_1}(x_1,r_1) \times B_{d_2}(x_2,r_2).$

We observe that for any $(w_1,w_2) \in B_d((y_1,y_2), r)$ ,

$\begin{aligned} d_i(w_i, x_i) &\leq d_i(w_i, y_i) + d_i(y_i, x_i) < r + d_i(y_i, x_i). \end{aligned}$

Hence, setting $\displaystyle r := \min_{i=1,2}\{r_i - d_i(y_i, x_i)\}$ yields the desired conclusion.

To prove $(\supseteq)$ , fix $B_d((x_1,x_2), r) \in \mathcal B_2$ and any element $(y_1, y_2)$ it contains. We seek $r_1 > 0 , r_2 > 0$ such that

$B_{d_1}(y_1,r_1) \times B_{d_2}(y_2,r_2) \subseteq B_d((x_1,x_2), r).$

We observe that for any $(w_1,w_2) \in B_{d_1}(y_1,r_1) \times B_{d_2}(y_2,r_2)$ ,

$\begin{aligned} d((w_1,w_2), (x_1,x_2)) &\leq d((w_1,w_2), (y_1,y_2)) + d((y_1,y_2), (x_1,x_2)) \\ &< r_1 + r_2 + d((y_1,y_2), (x_1,x_2)). \end{aligned}$

Hence, setting $r_1 = r_2 = \frac 12 (r - d( (y_1,y_2), (x_1,x_2) ) )$ yields the desired conclusion.

—Joel Kindiak, 28 Mar 25, 1538H

May 2, 2025