Does any set (with cardinality > 1) admit a topology that makes it a Polish space?

Solution 1:

First, let us discuss the case where $X$ is countable. Give $X$ the discrete metric (the distance between a point and itself is 0, the distance between any two distinct points is $1$), which induces the discrete topology on $X$. Now, note that any discrete metric space is complete: the only Cauchy sequences are those that are eventually constant (choose $\varepsilon < 1$), hence they converge. Furthermore, since $X$ is countable, $X$ is a countable dense subset of itself. So, $X$ is Polish.

The comment by mathcounterexamples.net addresses the case where $X$ is uncountable and has cardinality $\neq \mathfrak{c}$: since every uncountable Polish space has cardinality $\mathfrak{c},$ any uncountable set $X$ with cardinality not equal to $\mathfrak{c}$ cannot be given a topology that makes it a Polish space.

So, all that remains is the case where $X$ has cardinality $\mathfrak{c}.$ Clearly, there exists a bijection $f \colon X \to \mathbb{R}.$ We can define a topology on $X$ as follows: a subset $S \subseteq X$ is open if and only if $S = f^{-1}(U)$ for some open $U \subseteq \mathbb{R}$. (You should check that this actually does define a topology.) Furthermore, under this topology, $f$ automatically becomes a homeomorphism from $X$ to $\mathbb{R}.$ So, $X$ with this topology is homeomorphic to the Polish space $\mathbb{R}$, hence itself is Polish.

Now, to address some other parts of your general question:

• Given a set $X$, one can always give $X$ a topology that makes it compact: give it the indiscrete topology, where the only open sets are $\emptyset$ and $X$ itself. Since there are only finitely many open sets, any open cover is finite, hence a finite subcover of itself.
• Given a set $X$, one can always give $X$ a topology that makes it separable: once again, give it the indiscrete topology. Any singleton $\{x\} \subseteq X$ is contained in every nonempty open set of $X$ (there is only $X$ itself), hence every singleton $\{x\}$ is a countable dense subset of $X$.
• Given a set $X$, one can always give $X$ a topology that makes it metrizable: give it the discrete topology, where every subset of $X$ is open. Note that the discrete metric on $X$ defined above (the distance between a point and itself is 0, the distance between any two distinct points is $1$) induces the discrete topology, hence the $X$ with the discrete topology is metrizable.