Proving that equivalent metric spaces have the same closed set

Question

Given to metrics on $X$. $d_1$ and $d_2$ are said to be equivalent if for every $x\in X$ and $r>0$. $\exists \epsilon,\delta>0$

$$B_{d_1}(x,\epsilon)\subset B_{d_2}(x,r)$$

and $$B_{d_2}(x,\delta)\subset B_{d_1}(x,r)$$

Approach:

Since $d_1$ and $d_2$ are equivalent and that $G$ is open with respect to $d_1$. This means that for every $x\in G$, $\exists \epsilon>0$ such that $B_{d_1}(x,\epsilon) \subset G$. Since $d_1$ is equivalent to to $d_2$ we can find a $\delta>0$ such that $B_{d_2}(x,\delta)\subset B(x,\epsilon)$. Then for every $x\in G$ we have $B(x,\delta)\in G$ Using the same argument we can see that $G$ is open with respect to $d_2$

Questions:

How would I prove the conversley? (using that they are open sets then they have to be equivalent)

Would I use a similar approach to prove this for the closed set part?

Answer 1

This should be simple.

An interior point, $a$, of a set, $E$, with respect to a metric $d$ means there is an $r>0$ so that $B_{d}(a,r)\subset E$.

But if $D$ is an equivalent metric there will exists a $\delta>0$ so that $B_D(a,\delta)\subset B_{d}(a,r)\subset E$.

So an interior point of a set with respect to one metric will be an interior point of the set with respect to any equivalent metric.

A set is open in one metric if every point is an interior point. But if every point is an interior point in one metric, the every point will be an interior point in any equivalent metric. So the set will be open in all equivalent metrics.

And a set is closed if and only if its complement is open. So if the complement of a set is open with respect to one metric, the complement of the set will be open with respect to all equivalent metrics. SO if a set is closed in one metric, it is closed with respect to all equivalent metrics.

==== Addendum ====

It appears you are attempting to prove and if and only if statement.

So to prove that --if two metrics are such that a set open with respect to one metric will be open in respect to to the other, then the metrics are equivalent-- is pretty similar to the above.

For any $x\in X$ and $r> 0$ then $B_{d_i}(x,r)$ will be open in with respect to $d_i$.

But if $d_1$ and $d_2$ share the same open sets $B_{d_i}(x,r)$ will be be open with respect to both $d_1$ and $d_2$.

The definition of $B_{d_1}(x,r)$ being open in $d_2$ and the definition of $B_{d_2}(x,r)$ being open in $d_1$ mean there are $\delta, \epsilon > 0$ so that

$B_{d_2}(x, \epsilon)\subset B_{d_1}(x,r)$ and $B_{d_1}(x,\delta)\subset B_{d_2}(x,r)$

Answer 2

Let $T_i$ be the topology induced by $d_i$, $i = 1, 2$. We first show that $T_1 = T_2$ $\Longrightarrow$ $d_1$ and $d_2$ are equivalent.

Criterion
A set $U$ is open for $T_i$ if and only if $\forall x \in U, \exists \epsilon > 0,\, \, B_{d_i}(x, \epsilon) \subseteq U$.

Consider $x \in X$ and $r > 0$. Since $T_1 = T_2$, the open ball $B_{d_1}(x, r)$ is open set for $T_2$. hence, by the criterion above, there is $\delta > 0$ such that $B_{d_2}(x, \delta) \subseteq B_{d_1}(x, r)$.

Likewise, there is $\epsilon > 0$ such that $B_{d_1}(x, \epsilon) \subseteq B_{d_2}(x, r)$. QED

As for the question reguarding the closed sets : Since the colmplement of any open set is closed and vice versa,
the closed sets are the same $\Longleftrightarrow$ the open sets are the same.