I am interested in finding additional weak enough conditions for the following theorem to be true.
Let $X$ be a Polish space (that is, separable and completely metrizable), then if $\Omega \subset X$ is bounded, there exists a compact subset $K \subset X$ such that $\Omega \subset K$
Motivation: the theorem is by Heine-Borel true for finite dimensional real vector spaces (so $\mathbb{R}^d$'s in disguise) and my intuition tells me that if we impose separability, it is not too much of a stretch to go from a countable cover of some kind of extension of $\Omega$ (i.e. $\bar{\Omega}_{\epsilon}$ where $\Omega_{\epsilon}$ denotes the $\epsilon$ ball covering of $\Omega$ and bar denotes the closure) to a finite cover using some kind of neat trick or a weak additional assumption. I am trying to generalize a result I found in which $\Omega$ lies in $\mathbb{R}^{d}$ and in which this assumption of there being a great enough closed ball containing $\Omega$ is crucial for the rest of the argument.
Any hints, possible directions or other help would be greatly appreciated!
Best Answer
Your condition is, naturally enough, called the Heine-Borel property. It implies that $X$ is locally compact and $\sigma$-compact.