$f$ is a homeomorphism is equivalent to the following Prove the following theorem:

Let $f : (X, dx) \to (Y, dy)$ be one-to-one and onto. Then the following are equivalent:

*

*$f$ is a homeomorphism


*$x_n$ converges to $x$ with metric $d$ if and only if $f(x_n)$ converges to $f(x)$ with metric $d$


*$V$ is open in $X$ if and only if $f(V)$ is open in $Y$.

My attempt at an answer:

$1.$ and $2.$: if $f$ is a homeomorphism then $V = f
(−1)(U)$ is open in $X$ whenever $V$ is open in $Y$ (since $f$ is continuous), and $U = (f(−1))(−1)(V)$ is open in $Y$ whenever $U$ is open in $X$ (since $f(−1)$ is continuous). Conversely, if $2.$ holds
then both $f$ and $f(−1)$ have the property that preimages of open sets are open; so $f$ and $f(−1)$ are both continuous, and $1.$ holds.

I'm at a loss on how to prove the rest. Any help would be appreciated.
 A: $1\iff 3$: $f$ is a homeomorphism iff both $f$ and $f^{-1}$ are continuous iff $V\subset X$ is open if and only if $f(V)\subset Y$ is open (this is all just unpacking the definitions, really).
The interesting statement is the second one.
$1\implies 2$: Suppose $f$ is a homeomorphism. If $x_n$ converges to $x\in X$, then for any open ball $B_d(f(x))\subset Y$, there exists some $N$ such that for every $i>N$, we have $x_i\in f^{-1}(B_d(f(x)))$ (by definition of convergence, since $f^{-1}(B_d(f(x)))$ is open), and hence $f(x_i)\in B_d(f(x))$. Since this holds for any $d$, it follows that $f(x_n)\to f(x)$. A parallel argument shows the converse statement.
$2\implies 3$: Suppose $V\subset X$ is open. We aim to show $f(V)$ is open, or, equivalently, $f(V)^C$ is closed. Suppose not, and say there was a sequence $y_n\to y$ in $Y$ such that $y_n\in f(V)^C$, but $y\in f(V)$. Then, by (2), the sequence $f^{-1}(y_n)$, which is contained in $V^C$, would converge to $f^{-1}(y)\in V$, contradicting the fact that $V$ is open. Again, a parallel argument will show the converse and complete the proof.
