# Metrizability is a topological property?

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?

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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.

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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 $Y$.

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So how do I show that it generates the right topology? – Akaichan Mar 5 '13 at 6:04
On the other hand, I think I got it, thank you very much for your time. – Akaichan Mar 5 '13 at 6:10
@IvordesGreenleaf: You’re welcome. If you still have questions after you’ve thought about it more, feel free to ask. – Brian M. Scott Mar 5 '13 at 6:28

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

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