is union of nested compact spaces still compact?

Question

Stel $D$ a metric space. Let $K_1 \subset K_2 \subset K_3 \subset ...$ a serie of compact sets in $D$. I was wondering if $K = \bigcup_{n=1}^\infty K_n$ is compact too. If we take an open cover of $K$ then we can find for every $n$ a finite cover of $K_n$ but is there a way to extend to the case it is infinite?

Answer 1

Ascending chains like that will not preserve compactness, since the finite subcover for each level might get bigger and bigger.

As an example, look at $[-1,1] \subset [-2,2] \subset [-3,3] \dots$ in $\mathbb{R}$. Each set is compact, but the union is not.

Answer 2

If you want a finite interval example:

$$\;\left[\frac12,\,1\right]\subset\left[\frac13,\,1\right]\subset\ldots\subset\left[\frac1n,\,1\right]\subset\ldots$$

Each interval is closed and bounded and thus compact, yet their union is not:

$$\bigcup_{n=2}^\infty\left[\frac1n,\,1\right]=(0,1]$$

Answer 3

Take any infinite set, with the discrete topology, like a standard set of generators of $\ell_0$, and take an increasing family of finite subsets. Any member of the family is finite, thus it's compact. But the union is infinite and still discrete, so it can't be compact.

This is the smallest example I can come out with.