Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?
Here is my thought process.
If $X$ is a metric space then there exists some distance function on $X$ given by $d:X\times X\rightarrow\Bbb R$. Let $\varphi:X\rightarrow Y$ be a homeomorphism between $X$ and $Y$.
Then for any distinct points $y_1,y_2\in Y$ we can determine a "distance" between them by $d(\varphi^{-1}(y_1),\varphi^{-1}(y_2))\in\Bbb R$
To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y_1,y_2) = d(\varphi^{-1}(y_1),\varphi^{-1}(y_2))$?
If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?
Best Answer
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $\phi: X \to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $\psi: Y \to X$ so that $\psi \circ \phi = 1_X$ and $\phi \circ \psi = 1_Y$), and we can define $$d_Y: Y \times Y \to \mathbb{R} \text{ by: } d_Y(y_1,y_2) = d_X(\psi(y_1), \psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $\phi[B_{d_X}(x, r)] = B_{d_Y}(\phi(x), r)$ for all $x \in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.