It might be ridiculous but I cannot comprehend why only closed subsets of compact sets are compact. I know there are easy counter examples like $(0,1) \subseteq [0,1]$ where $(0,1)$ is not compact but $[0,1]$ is compact. However, I am still confused.

Let $K$ is a compact set in topological space. Then for ever open cover, there exist some finite subcover.

If $\bigcup_{ \alpha \in I} O_{\alpha}$ is an open cover for $K$ there exists $\bigcup_{ \alpha \in J} O_{\alpha}$ is a finite open cover for $K$ with $J\subseteq I$

Let $A$ is any subset of $K$.

Why I cannot say $A \subseteq K \subseteq \bigcup_{ \alpha \in J} O_{\alpha}$ ? Why $\bigcup_{ \alpha \in J} O_{\alpha}$ doesn’t cover $A$?

I need help for getting rid of this confusion. Thanks in advance

  • 4
    $\begingroup$ Not every open cover for $A$ is an open cover for $K$ $\endgroup$
    – clark
    Dec 29 '17 at 20:33
  • $\begingroup$ @clark every open cover for $K$ is an open cover for $A$, isn’t it? $\endgroup$
    – usereb
    Dec 29 '17 at 20:35
  • 1
    $\begingroup$ @esrabasar Yes, but that's not what clark said: they're talking about the other direction. This is spelled out more in my answer. $\endgroup$ Dec 29 '17 at 20:36
  • $\begingroup$ Just some thoughts. If that would be the case then on $\mathbb{R}$ every bounded subset would be compact since it is in some interval $[-K,K]$ which is compact. Why bother introducing compactness then, if it is just boundedness? I understand your misunderstanding btw $\endgroup$
    – Shashi
    Dec 29 '17 at 20:42

You have correctly observed that

Every cover of $K$ is also a cover of $A$.

However, the key point here is

There are covers of $A$ which are not covers of $K$.

Note the changing order of $A$ and $K$, in the above two sentences.

For example, let's look at $(0, 1)$ versus $[0, 1]$. We can tell that $(0, 1)$ isn't compact by thinking about the open cover $$\{(0, 1-{1\over n}): n\in\mathbb{N}\}$$ which is infinite but has no finite subcover. However, note that this doesn't cover $[0, 1]$. So the compactness of $[0, 1]$ doesn't give us any information here.


Let $A$ be compact and let $B\subset A$ not be compact. Then it is certainly true that if $\{O_i\}$ is an open cover of $A$ it is also an open cover of $B$, and some finite subset is an open cover of $B$. What you’re not thinking about is that by shrinking the set under consideration, you allow for more collections of sets to be open covers. There are collections of set that are open covers of $B$ that are not open covers of $A$, and those collections are the ones without finite subcovers.

To put this concretely, let’s look at your example of $[0,1]$ and $(0,1)$. If we define $K=\{(1/n,1-1/n):n\in \mathbb N\}$ we find that no finite subcover of $K$ will cover $(0,1)$. So why doesn’t this contradict the compactness of $[0,1]$? $K$ itself is not a cover of $[0,1]$, so the compactness of $[0,1]$ doesn’t tell us anything about $K$.


We should go back and look at the definition of a compact set: A set $A$ is compact if any open cover of $A$ there exists a finite sub cover. Thus, you should look at the open cover of $A$ not $K$. For instance, $(0,1-\frac{1}{n})$ is an open cover of $(0,1)$ but not $(0,1]$, you cannot apply your argument.


Of course the $O_\alpha$'s cover $A$. That't not the problem. The problem is that not all open covers of $A$ can be obtained by this process. For instance, in the case of $(0,1)$, consider the open cover$$\left\{\left(\frac1n,1-\frac1n\right)\,\middle|\,n\in\mathbb{N}\setminus\{1,2\}\right\},$$which has no finite subcover.


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