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I have a doubt at the definition of compact spaces. So if you have a topological space $X$, then $X$ is compact if every open cover of $X$ has a finite subcover. In other words, if $X$ is the union of a family of open sets, there is a finite subfamily whose union is $X$.

So, if $\{A_i\}$ is a family of open sets such that it is a finite subcover of $X$ so $X=$ union of all $A_i$. Ok, until here I got it. But now if I consider that there is an $j$ such that $A_j=X$, which is open, so of course that we have ALWAYS: $X$=union of all $A_i$.

My doubt: this might be stupid but if I can do this always so the set is always compact...??

Thanks in advance to the ones who help me :D :D

(Source of definition: http://mathworld.wolfram.com/CompactSpace.html)

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    $\begingroup$ it has to be the case that every open cover has a finite sub-cover, not just that there exists at least one. So if someone gives you an open cover, that open cover may not contain an open set equal to the space. $\endgroup$ – Thoth Dec 11 '15 at 0:27
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    $\begingroup$ Why is there a $j$ such that $A_j=X$? $\endgroup$ – Eric Wofsey Dec 11 '15 at 0:27
  • $\begingroup$ You can write things like $X=\bigcup A_i$, $X=\bigcup_{i\in I} A_i$, $X=\bigcup\limits_{i\in I} A_i$. Just enter: $X=\bigcup A_i$, $X=\bigcup_{i\in I} A_i$, $X=\bigcup\limits_{i\in I} A_i$. For more information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Dec 11 '15 at 4:05
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No, the definition says:

For every cover, there exists a finite subcover.

This means that you cannot select the sets $A_j$. You are given a set $X$, and a set of sets $\{A_j\}_{j\in J}$. Then, no matter what the sets $A_j$ are (you do not know if one of them is equal to $X$), you can find a finite set $A_1,A_2,\dots, A_n$ such that they cover $X$.

Sure, there may be (and in fact always are) covers of $X$ which do contain finite subcovers. A typical example of this is taking any cover that contains $X$ as one of its covering elements, as then $\{X\}$ is obviously a finite subcover. But that's not the point. The point is that there must be a finite subcover no matter what the original cover was.

For example, $X=\mathbb R$ is not compact, because I can define

$$A_i = (i-1, i+1)$$

and then $\{A_i, i\in \mathbb Z\}$ becomes a cover for $\mathbb R$. However, there is no finite subcover that covers $\mathbb R$.

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