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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?

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  • $\begingroup$ A locally compact separable metric space is the union of an ascending chain of compact subsets, each one contained in the interior of the next. $\endgroup$
    – egreg
    Apr 4, 2015 at 10:08

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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.

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  • $\begingroup$ Will descending chains? $\endgroup$ May 13, 2020 at 22:30
  • $\begingroup$ @iaskdumbstuff in Hausdorff contexts (such as metric spaces), the intersection of a descending chain of compact spaces will be closed (as compact spaces are closed) and closed subsets of compact sets are compact. In non-Hausdorff context my intuition is currently failing me $\endgroup$
    – Rolf Hoyer
    May 13, 2020 at 23:08
  • $\begingroup$ Thanks for your reply. What about infinite unions of descending chains (in Hausdorff contexts)? The usual counterexample to prove that an infinite union of compact sets is not necessarily compact is for the sets $A_n = [-n, n]$, but this set wouldn't be a part of a descending chain. Since the sets in a descending chain cannot get "bigger and bigger", I'm having trouble finding a counterexample. $\endgroup$ May 13, 2020 at 23:15
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    $\begingroup$ @iaskdumbstuff infinite unions of anything are unlikely to be compact, I would say. For instance, if you had a countable sequence of descending chains with each chain having intersection a single point, their union could be assembled to give rise to any countable set, which need not be compact $\endgroup$
    – Rolf Hoyer
    May 13, 2020 at 23:38
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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:

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

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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.

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