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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?

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    $\begingroup$ I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space. $\endgroup$ – Lord Shark the Unknown Sep 11 '18 at 21:03
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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.

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You construction can be done regardless of $\varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(\varphi^{-1}(-),\varphi^{-1}(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y \in Y$ and $\varepsilon > 0$ . Now,

$$ B'_\varepsilon(y) := \{x:d'(x,y) < \varepsilon\} = \{x : d(\varphi^{-1}(x), \varphi^{-1}(y)) < \varepsilon\} = \varphi(B_\varepsilon(\varphi^{-1}(y))). $$

Since $B_\varepsilon(\varphi^{-1}(y))$ is a ball in $X$ and $\varphi$ is homeo, $B'_\varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U \subseteq Y$ is open for the original topology,

$$ U = \varphi(\varphi^{-1}(U)) = \varphi(\bigcup_{i \in I}B_{r_i}(x_i)) = \bigcup_{i \in I}\varphi(B_{r_i}(x_i)) = \bigcup_{i \in I}B'_{r_i}(\phi(x_i)) $$

and so $U$ is open for the metric topology. Here we use that $\varphi^{-1}(U)$ is an open set of $X$ and so it is a union of open balls.

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I would say is as simple as that, since $\varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.

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Given any sets A and B and a bijection $\varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $\varphi(\varphi^{-1}(b_1)+\varphi^{-1}(b_2))$, then $\varphi$ will be a group isomorphism with respect to addition in A and addition in B.

Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $\varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.

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