KindiakMath

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).

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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.

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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.

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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.

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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

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