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Is metrizability preserved under homeomorphism? That is, suppose that you have a topological space $(X, \tau_1)$ whose topology comes from a metric $d$, and you have another topological space $(Y, \tau_2)$ with $Y$ homeomorphic to $X$. Can you find a metric on $Y$ that is induced by the topology $\tau_2$?

Let $h:Y \to X$ be a homeomorphism, where $(X,d)$ is a metric space. Then the function $D: Y\times Y \to R$ given by $D(a,b) = d(h(a),h(b))$ is a metric on $Y$ that induces its topology. Is that true? I have trouble verifying it.

Let $T$ be the topology on $Y$, and let $S$ be the topology on $Y$ induced by $D$. Let $U$ be a member of $T$ and let $y$ be in $U$. Then $h(U)$ is open in $X$ and contains $h(y)$. So there exists an open ball $B_d(h(y),e)$ contained in $h(U)$. Then $h^{-1}(B_d(h(y),e))$ is an open neighbourhood of $y$. But where is the $D$-open ball about $y$ contained in $U$? Is it $B_D(y,e)$???

mattosmat
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user130390
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    If $\eta:Y\to X$ is a homeomorphism and $\rho$ is a metric on $X$ for $\tau_1$, consider $\rho':Y\times Y\to\Bbb R$ given by $\rho'(x,y)=\rho(\eta x,\eta y)$? – Pedro Feb 21 '14 at 02:34

2 Answers2

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Since $ h:Y\rightarrow X $ is a homeomorphism where $(X,d)$ is a metric space and $(Y, \tau)$ is a topological space, the fuction $ \rho:Y\times Y\rightarrow [0,\infty) $ given by $ \rho(y_{1},y_{2})=d(h(y_{1}),h(y_{2})) $, for all $ y_{1},y_{2}\in Y $ is a metric on Y. Let $A\in \tau$. Since $ h^{-1} $ is a continuous function $h(A)$ is d-open in X. Let $y\in A$. Then $h(y)\in h(A)$ and there is an open ball $ B_{d}(h(y),r) $ such that $ h(y)\in B_{d}(h(y),r)\subseteq h(A) $. Since h is a continuous function and $ \rho(y_{1},y_{2})=d(h(y_{1}),h(y_{2})) $, we have $ h^{-1}(B_{d}(h(y),r))=B_{\rho}(y,r) $ and $ y\in B_{\rho}(y,r)\subseteq A $. Therefore A is $\rho$-open subset of Y and hence $\tau \subseteq $ {U: $U\subseteq Y$ and $U$ is $\rho$-open}. Similarly we can show that {U: $U\subseteq Y$ and $U$ is $\rho$-open} $\subseteq$ $\tau$. Therefore $\tau = $ {U: $U\subseteq Y$ and $U$ is $\rho$-open}. So $(Y, \tau)$ is metrizable and $\rho$ is the metrization of $\tau$.

ASB
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Homeomorphisms preserve topology, which is the only structure a general topological space has. Therefore, they preserve every property that can be defined in a general topological space. Metrizability is one such property.

If the preceding is not convincing, consider the following: a topological space is metrizable iff it is homeomorphic to some metric space. Since the composition of homeomorphisms is a homeomorphism, the claim follows.

user127096
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