Exercise 21.1

Proof. Clearly $d{\text{|}}_{A\times A}$ is a metric since it inherits all the properties for a metric from the metric $d$ for $X$; it therefore suffices to show that every basis element for the subspace topology on $A$ contains some open ball defined by $d{\text{|}}_{A\times A}$, and vice versa, by Lemma $13.3$.

So, suppose $B$ is a basis element for the metric topology on $A$; $B=B_{d{\text{|}}_{A\times A}}(x,r)$ for some $x\in A$ and $r\in ℝ$. But then, $B=B_d(x,r)\cap A$ by definition, and so the subspace topology is finer than the metric topology.

Conversely, suppose $B$ is a basis element for the subspace topology on $A$; it equals $B_d(x,r)\cap A$ for some basis element $B_d(x,r)$ of $X$. Let $y\in B_d(x,r)\cap A$; we see that the set $B_{d{\text{|}}_{A\times A}}(y,r-d(x,y))\subseteq A\cap B_d(x,r)$ is a basis element for the metric topology contained in $A\cap B_d(x,r)$, since $z\in B_{d{\text{|}}_{A\times A}}(y,r-d(x,y))$ is such that

Thus, the metric topology is finer than the subspace topology. Combining the two inclusions, we see the topologies are equal. □

Proof. Let us denote the restricted function $d\upharpoonright A\times A$ by $d^{\prime }$. Clearly $d^{\prime }$ is a metric on $A$, since for $x,y\in A$ we have $d^{\prime }(x,y)=d(x,y)$ and $d$ has all the properties of a metric. We show that the metric topology induced by $d^{\prime }$ is the same as the subspace topology using Lemma 13.3 in both directions.

First we show that $B_{d^{\prime }}(x,𝜖)=A\cap B_d(x,𝜖)$ for any $x\in A$ and $𝜖>0$. For any $y\in B_{d^{\prime }}(x,𝜖)$ we have that clearly $y\in A$ since the metric $d^{\prime }(y,x)$ must be defined, and $d(y,x)=d^{\prime }(y,x)<𝜖$ so that also $y\in B_d(x,𝜖)$. Thus $y\in A\cap B_d(x,𝜖)$, and hence $B_{d^{\prime }}(x,𝜖)\subset A\cap B_d(x,𝜖)$. For the other direction suppose that $y\in A\cap B_d(x,𝜖)$ so that $y\in A$ and $y\in B_d(x,𝜖)$. Thus $y$ and $x$ are in $A$ so that $d^{\prime }(y,x)$ is defined and $d^{\prime }(y,x)=d(y,x)<𝜖$, and hence $y\in B_{d^{\prime }}(x,𝜖)$. This shows that $B_{d^{\prime }}(x,𝜖)\supset A\cap B_d(x,𝜖)$, which in turn shows that the two sets are the same.

Now, clearly, each basis element of the metric topology $B_{d^{\prime }}(x,𝜖)=A\cap B_d(x,𝜖)$ is also a basis element of the subspace topology by Lemma 16.1. Hence for any $x\in A$ we have that $B_m=B_{d^{\prime }}(x,𝜖)$ is a basis element of the metric topology and $B_s=A\cap B_d(x,𝜖)$ is a basis element of the subspace topology. But $B_m=B_s$ so that $x\in B_s\subset B_m$ and $x\in B_m\subset B_s$ so that each topology is finer than the other by Lemma 13.3. Thus they are the same topologies. □