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Let $d_1$ and $d_2$ be two equivalent metrics. If $A$ is an open subset of $(X,d_1)$, is $A$ an open subset of $(X,d_2)$?

Well, I have that if two metrics are equivalent then every sequence in $(X,d_1)$ is convergent in $(X,d_2)$. Now, let $a_n$ be an arbitrary sequence that converges to $x_0in A^c subset X$ under $d_1$, with $x_0$ a limit point (or accumulation point). Then, the sequence converges by definition of equivalent metrics in $(X,d_2)$. Now, we know that $a_n$ converges to some point $x_1$ in $(X,d_2)$ but not necessarily $x_0$ = $x_1$ (I can prove that). So then, it can be that in $A^c subset X$ under $d_2$ does not contain every accumulation point (I’m saying that it can be $x_1 notin A^c$). Then $A^c$ isn’t a closed set and we cannot prove anything. Or until now I can’t find a way to prove that $x_1in A^c$.

Any suggestions? Am I right?

A time ago I proved in some homework that if some metrics $d_1$ and $d_2$ are equivalent then we can bound each other, something like:

$$exists C : d_1 leq C d_2 land exists K : d_2 leq K d_1$$

But I can’t prove it to the other side, it may help

Let $A subseteq X$ be open with respect to $d_1$. Because $d_1$ and $d_2$ are equivalent, there exists $C_1,C_2 > 0$ such that
$$
C_1 d_1(x,y) leq d_2(x,y) leq C_2d_1(x,y)
$$
for all $x,y in X$. Let now $x in A$ be given. Since $A$ is open with respect to $d_1$, there exists $epsilon > 0$ such that
$$
B(x,epsilon; d_1) subseteq A.
$$
Here, $B(x,epsilon; d_1)$ denotes the open ball (with respect to $d_1$)
$$
left y in X : d_1(x,y) < epsilon right.
$$
Now, consider the open ball (with respect to $d_2$):
$$
B := Bleft( x, C_1epsilon; d_2right).
$$
We claim that $B subseteq B(x,epsilon; d_1) subseteq A$ which would complete the proof. Indeed, if $y in B$ then
beginalign*
d_1(x,y)leq fracd_2(x,y)C_1 < fracC_1epsilonC_1 = epsilon.
endalign*
Hence, $y in B(x,epsilon;d_1) subseteq A$. By symmetry, you can then conclude that open subsets of $(X,d_2)$ are open in $(X,d_1)$ as well. This means that equivalent metrics induce the same topology on $X$.

Edit: Let's instead only assume that $d_1$ and $d_2$ satisfy the following property:

A sequence $(x_n)$ converges to $x in X$ with respect to $d_1$ if and only if it converges to $x$ with respect to $d_2$.

If $A$ is open with respect to $d_1$, then it will still be open with respect to $d_2$. To verify this, it would be enough to check that $A^complement$ is closed with respect to $d_2$. To this end, let $(x_n)$ be a sequence in $A^complement$ converging to $x in X$ with respect to $d_2$. Then, we have $x_n to x$ in $(X,d_1)$ by assumption. Since $A^complement$ is closed in $(X,d_1)$, we get that $x in A^complement$. This ensures that $A^complement$ is closed in $(X,d_2)$, and so $A$ is open in this same space.