Metrizability is a topological property?

Question

How could I show that metrizability is a topological property?

Well, this means that if I have a set $X$ that is metrizable and a homeomorphic function $f$ from $X$ to $Y$, then I need to show that $Y$ is metrizabke, correct?

If I let $d$ be a metric in $X$? How do I construct a metric in $Y$ using the bijection $f$?

Answer 1

Hint: homeomorphism transport the structure of $X$ to the one in $Y$. Try to use that fact to "pull" the metric $d$ to some metric $\tilde d$ on $Y$ such that $f$ is now an isometry.

Answer 2

Hint: Try letting $d_Y(a,b)=d(f^{-1}(a),f^{-1}(b))$.

Answer 3

Just let $\rho(y_1,y_2)=d\big(f^{-1}(y_1),f^{-1}(y_2)\big)$ and check (1) that $\rho$ is a metric on $Y$, and (2) that it generates the right topology.

Note that this should be the obvious thing to try. The fact that $X$ and $Y$ are homeomorphic means that from a topological point of view $Y$ is just $X$ under a different name, and the homeomorphism $f$ is the ‘translator’ from $X$ to $Y$. Thus, $f(x)$ should behave in $Y$ exactly as $x$ does in $X$, and if we can fit a certain distance $d(x_1,x_2)$ to two points of $X$ in a way that fits the topology of $X$, we ought to be able to fit the same distance between the points $f(x_1)$ and $f(x_2)$ in a way that fits the topology of $Y$.

Answer 4

We can use the Nagata- Smirnov metrization theorem.