I was recently trying to verify certain things in the setting of Locally compact 2nd countable Hausdorff spaces. I thought that this is a natural collection of spaces more general than metric spaces, but I think that following the metrization theorems that such spaces must be metrizable with a complete metric. I didn't find such an explicit statement, so I thought to check whether this is indeed correct.
My reasoning\intuition is as follows.
- A locally compact Hausdorff space is completely regular.
- By Uryshonn's metrization theorem, a completely regular 2nd countable space is metrizable.
- Every 2nd countable locally compact Hausdorff space is also $\sigma$-compact.
- For such a space, closure of balls should be compact.
The last point is imprecise, but I have the strong suspicion that such a statement must hold. I assumed if I am wrong someone can generate a well known counter-example. Also if this statement is correct, this should be known under some terminology which I don't know how to look for.
I would appreciate any directions on this.
Best Answer
It's certainly not true that in a second-countable locally compact metric space, closures of balls are compact. For instance, consider $(0,1)$ with the usual metric.
Instead, you have to find a way to construct a special metric for which this is true. Here is one way to do so. Let your space be $X$ and let $X^*=X\cup\{\infty\}$ be its $1$-point compactification. If $(U_n)$ is a countable basis for $X$ consisting of sets with compact closures, then there is a countable local base at $\infty$ in $X^*$ consisting of the complements of the closures of finite unions of the $U_n$'s (since a neighborhood of $\infty$ is the complement of a compact set in $X$ which must be covered by finitely many of the $U_n$'s). Combining this countable local base with $(U_n)$, you get a countable basis for $X^*$, so $X^*$ is also metrizable. Now take a continuous function $f:X^*\to\mathbb{R}$ that vanishes only at $\infty$ (e.g., $f(x)=d(x,\infty)$ for some metric $d$) and embed $X$ into $X\times\mathbb{R}$ by $x\mapsto(x,1/f(x))$. This gives a metric on $X$ with the property that any bounded set has compact closure (since a set where $1/f$ is bounded is contained in the complement of a neighborhood of $\infty$ in $X^*$).