Topologically equivalent metrics using sequence

Question

Now I am studying about topologically equivalent of two metrics on General Topology. There is an exercise:

Given $d$ and $\rho$ are metrics on X which topologically equivalent if only if for an arbitrary $(x_n)\subseteq X$ then $d(x_n,x)\rightarrow 0 \Longleftrightarrow \rho(x_n,x)\rightarrow 0$

I try to prove that problem,

$(\Rightarrow)$ Let $U \subseteq X$ be an open set on $(X,d)$. Thus, for all $x_n \in U$ there is $\epsilon > 0$ such that $$B_d (x_n,\epsilon) \subseteq U$$ Because $B_d(x_n,\epsilon)$ is open on $(X,d)$ and by hypothesis, $d$ and $\rho$ are equivalent, so $B_d(x_n,\epsilon)$ is also open on $(X,\rho)$. Thus, for all $(x_n)$ there is $\epsilon^\ast$ such that $$B_\rho (x_n,\epsilon^\ast) \subseteq B_d (x_n,\epsilon)$$ This means that there is a point $x$ such that $\rho(x_n,x)<\epsilon^\ast$ and $d(x_n,x)<\epsilon$. Thus, $d(x_n,x)\rightarrow 0 \Rightarrow \rho(x_n,x)\rightarrow 0$. Similarly, can be obtained $d(x_n,x)\rightarrow 0 \Leftarrow \rho(x_n,x)\rightarrow 0$.

Thus, $d(x_n,x)\rightarrow 0 \Longleftrightarrow \rho(x_n,x)\rightarrow 0$.

$(\Leftarrow)$ Let $(x_n)\subseteq X$. By hypothesis, let $\epsilon >0$ and $x\in X$ then $d(x_n,x)<\epsilon$ implies $\rho(x_n,x)<\epsilon$, this means that $B_\rho(x_n,\epsilon)\subseteq B_d(x_n,\epsilon)$. Thus, $\tau_\rho \subseteq \tau_d$. Similarly, can be obtained $\tau_d \subseteq \tau_\rho$.

Thus, $\tau_d = \tau_\rho$ and this means that $d$ and $\rho$ are equivalent.

But, I am not sure with my work. Is there any wrong steps in my work? Any correction will be appreciated.

Answer 1

Assume that $d$ and $\rho$ obey the "sequence property".

Lemma (classical fact):

If (X,d) is any metric space and $(x_n)_n$ is a sequence in $X$ and $x \in X$, then $x_n \to x$ (in the topology on $(X,d)$) iff $d(x_n,x) \to 0$ (in $\Bbb R$).

Let $A$ be closed in $(X,d)$. Suppose that $A$ were not closed in $(X,\rho)$ then there would be a point $x\in X\setminus A$ and a sequence $a_n \in A, n \in \Bbb N$ such that $x_n \to x$ under $\rho$. The latter means that $\rho(a_n, x) \to 0$ by the lemma and by the sequence property we have that $d(a_n,x) \to 0$ and so applying the lemma again in reverse direction we get that $a_n \to x$ in $(X,d)$ and as $A$ is closed in $(X,d)$ it follows that $x \in A$, contradiction. So $A$ is closed in $(X,\rho)$ too.

A symmetric argument gives us that a closed set in $(X,\rho)$ is also closed in $(X,d)$ so these metric spaces have the same closed sets and thus the same open sets and so are equivalent.

Now, assume the metric spaces are equivalent and we show the sequence property. So let $x_n$ be a sequence in $X$ and $x \in X$ and suppose that $d(x_n,x) \to 0$. This means that $x_n \to x$ in $(X,d)$ (lemma) and as $(Y,\rho)$ is equivalent we have $x_n \to x$ in $(Y,\rho)$ as well (convergence depends only on he open sets and these are the same) and so $\rho(x_n,x) \to 0$ again (lemma). The reverse implication is entirely similar. Summarised:

$$d(x_n,x) \to 0 \iff x_n \to x \text{ in } (X,d) \iff x_n \to x \text{ in } (X,\rho) \iff \rho(x_n,x) \to 0$$