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I don't understand how a compact set can have a finite open subcover given certain infinite open covers.

For example, say my set $K\subset \mathbb{R}^n$ is the closure of some ball around a point: $K=\overline{B_r(p)}$. Since we're in Euclidean space and $K$ is closed and bounded, $K$ is compact. Consider the sequence of open sets $\{\overline{B_{r/n}(p)}^c\}_{n=1}^\infty$. Their infinite union is an open cover of $K$ but any finite union isn't; namely, any finite union doesn't contain $p$.

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    $\begingroup$ Your union does not cover $p$. $\endgroup$
    – Son Gohan
    Commented Jan 1, 2022 at 21:41
  • $\begingroup$ In fact, none of your sets contain $p$. $\endgroup$
    – John Douma
    Commented Jan 1, 2022 at 21:44
  • $\begingroup$ That's not a cover of $K$. The union doesn't contains the whole $K$ $\endgroup$
    – jjagmath
    Commented Jan 1, 2022 at 21:44
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    $\begingroup$ And a clarification about the way you are using the term "open cover". An open cover of $A$ is a set of open sets whose union contains $A$. You used the term "open cover" to refer to the union, which is not right. $\endgroup$
    – jjagmath
    Commented Jan 1, 2022 at 21:50
  • $\begingroup$ @jjagmath that's fair - imprecise wording on my part. Though I suppose to be pedantic the union is also an open cover singleton. $\endgroup$ Commented Jan 1, 2022 at 21:55

2 Answers 2

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$p$ isn't in $\cup_{n=1}^\infty \overline{B_{r/n}(p)}^c,$ so it's not an open cover:

$p=\cap_{n=1}^\infty \overline{B_{r/n}(p)}=\left(\cup_{n=1}^\infty \overline{B_{r/n}(p)}^c\right)^c$

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I see your point but note that $\{\overline{(B_{r/n}(p))}^{c}\}_{n=1}^\infty$ is not covering the center $p$. Their infinite union is not an open cover of $K$, since the point $p$ would always be left out from the union.

To see why, consider the unidimensional example of this sequence of intervals $(1/n, 1)$. If you take the countable union among all naturals you are not going to cover the interval $[0,1)$ since in the union you will get $(0,1)$. Formally this is explained by saying that a countable union of open set is open, hence it must be $(0,1)$ and not $[0,1)$ since the latter is not open.

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