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I've read brief topology part of my analysis book and came across the page of concept called compactness.

I've understood at least the definition of a set being compact (One shows $K$ is compact subset of space $X$ if for every open cover of $K$ there must be subcover of it which is finitely indexed), could prove some propositions like:

"Any Compact set of metric spaces are closed",

"If any finite subcollections of collection of compact sets $\{K_q\}$ is nonempty then intersection of all its elements is nonempty".

Thought I was going in right way and however, I really wasn't at all. I know there are various proofs that tells $[0,1]$ is compact set and they are rigorously right.

But in order hand my intuition tells that :

for any $N$ positive integer, $\{(-\frac1N,\frac1N),(0,\frac2N),(\frac1N,\frac3N),...,(1-\frac1N,1+\frac1N)\}$ covers our interval fine and it is infinite but one really can't expect any subset of this to be covering $[0,1]$.

If we remove $(\frac kN,k+\frac 2N)$ (where $k=0,1,\dots ,n-1$) from the list, it clearly won't cover point $k+\frac 1N$ of $[0,1]$ and no need to talk about when it is reduced to finite size. What am I missing here?

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  • $\begingroup$ Hi! This looks like a thoughtful and interesting question, but unfortunately, your notation is very difficult to read. Please learn some MathJax to make this question (and your future questions) more legible, and in keeping with the site's standards. $\endgroup$ – Theo Bendit Oct 11 '18 at 11:13
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    $\begingroup$ Your cover is already finite; it has $N$ elements. $\endgroup$ – hartkp Oct 11 '18 at 11:14
  • $\begingroup$ Well thanks for giving me comment hartkp but i was considering N to diverge to infinity and what would happen. $\endgroup$ – Bosark Nreuifedog Oct 11 '18 at 11:17
  • $\begingroup$ @BosarkNreuifedog What would happen is that you'd get a sequence of open covers, all of which are finite, but that grow unboundedly in number. $\endgroup$ – Theo Bendit Oct 11 '18 at 11:21
  • $\begingroup$ @TheoBendit. N is constant in each of those terms. $\endgroup$ – William Elliot Oct 11 '18 at 11:54

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