# Example of a locally compact metric space which is $\sigma$-compact but not proper

Let $(X,d)$ be a locally compact metric space. Then it is known that $X$ is separable if and only if it is $\sigma$-compact (i.e. it can be written as a countable union of compact sets). Moreover, under this condition there is a proper function $f:X\to\mathbb{R}$ and $$d'(x,y)=d(x,y)+|f(x)-f(y)|$$ is a compatible metric (i.e. it generates the same topology) sucht that $(X,d')$ is proper (i.e. every closed ball is compact).

Now here is my question: Is there an example of a locally compact metric space which is $\sigma$-compact but not proper?

• If "every closed ball is compact" is the definition of proper then the interval $(0,1)$ with the usual metric is a $\sigma$-compact metrix space which is not proper. – David C. Ullrich May 25 '16 at 14:22
• I don't understand: if it is $\sigma$-compact it is separable etc. so we have the proper function? – Henno Brandsma May 25 '16 at 14:22
• @HennoBrandsma That proper function shows that $(X,d')$ is proper, while the question, I think, is whether $(X,d)$ must be proper... – David C. Ullrich May 25 '16 at 14:23
• @David C. Ullrich I accept your example. I was obviously thinking in the wrong direction. Thanks! – gerald May 25 '16 at 15:05

Here is an example with one more property: $X$ is complete. Namely, let $X$ be the real line with the metric $d(x,y)=\min(|x-y|,1)$. At small scales this is the same as Euclidean metric, so the space is complete and locally compact. It is also $\sigma$-compact, as the union of intervals $[n,n+1]$.
On the other hand, it is not proper since any closed ball of radius $1$ is the entire space, which contains the infinite subset $\mathbb{Z}$ without a limit point.