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I'm interested in one question about Heine-Borel theorem. We know that if S is bounded and closed then it's compact. Standard proof using fact that some segment is compact. And because of S is bounded it's have left and right bound of segment. But my question is my S should be closed? We could take open , bounded S and some segment which cover out set. Why it's so necessary ?

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    The question is not clear. Are you asking what the role of closedness is in the proof of HB theorem, or are you asking why a set needs to be closed in order for it to be compact? If the latter, you should phrase your question as follows: "why is every compact set necessarily a closed set?". If the former, you should read the proof again. – 2017-01-07
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    Your space should also satisfy bolzano wiestrass theorem....so that every bounded sequence has a convergent subsequence and since the set set is closed the limit will belong to the set – 2017-01-07

3 Answers 3

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Let $S=(0,1)$ and consider the cover $\{(1/n,1) : n \ge 2\}$. The union of any finite subset of the cover is of the form $(1/n,1)$ for some $n$, so no finite subset can cover $S$.

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Let $S$ be any subset of $\Bbb R$ that is not closed; then $S$ is not compact.

To see this, note first that since $S$ is not closed, it has a limit point $p$ that is not in $S$. For each $n\in\Bbb Z^+$ let

$$U_n=\left\{x\in\Bbb R:|x-p|>\frac1n\right\}\;;$$

each $U_n$ is an open set in $\Bbb R$, and $\bigcup_{n\in\Bbb N}U_n=\Bbb R\setminus\{p\}\supseteq S$, so $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ is an open cover of $S$. Suppose that $F$ is a finite subset of $\Bbb Z^+$. Let $m=\max F$; then each $U_n$ with $n\in F$ is a subset of $U_m$, so $\bigcup_{n\in F}U_n=U_m$. But $p$ is a limit point of $S$, so there is an $x\in S$ such that $|p-x|<\frac1m$, and clearly $x\in S\setminus U_m$. Thus, no finite subfamily of $\mathscr{U}$ covers $S$.

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An open interval, say $(0,2)$ is not compact in the reals, while $[0,2]$ is.

Generally a subset of a compact set is not compact.

Note that while one characterization of compact is that every sequence in $C$ has a convergent subsequence this means it has to converge in $C$, that is the limit must be in $C$.

So for example the sequence $(1/n)_{n\ge 1}$ shows that the set $(0,2)$ is not compact. The sequence does not converge as a sequence in $(0,2)$ since its limit (in the reals) is outside the set.

If you consider the definition via open coverings the point is that a smaller will have coverings that won't covert the larger and those might not allow a finite subcover.

Also this may be counter-intuitive at first, but covering the larger sets might give you a "good" building block for a cover that allows you to ditch many other sets from a cover of the smaller set.

Take the example basically as in the other answer $(1/n,2)$ for $n\ge 1$ which is an open cover of $(0,1]$ with no finite subcover. But if you were to cover the point $0$ too (so the compact interval $[0,1]$), you'd need a "new" open set and this open set will allow you to discard almost all of the former sets. Say you cover $0$ with $(-0.001, 0.001)$ then you only need the former sets for $n$ up to $1001$ to cover $[0,1]$, so you have a finite cover. If you cover $0$ with something still smaller just increase the $n$ but always a finite number suffices as you'll cover $0$ with an open set that thus contains some interval $(-\epsilon, \epsilon)$.