1. An open covering of a space $X$ is a collection $\{U_i\}$ of open sets with $\displaystyle X=\bigcup_{i\in I}U_i$.

  2. If $X$ is compact, for every open cover there exist at least a finite subcover, so this happens $X\displaystyle= \bigcup_{i=1}^nU_i.$

However with 1. and 2. aren't we saying that $$\displaystyle \bigcup_{i\in I}U_i= \bigcup_{i=1}^nU_i$$


But this means that the union of open sets (maybe not finite) is equal to the union of finite open sets.

Why does that happen?

  • 3
    $\begingroup$ You are correct that compactness has to do with finiteness. I think the answers at this thread do a good job of explaining the concept. $\endgroup$ – Adam Lowrance Dec 17 '17 at 2:30

Beware that you can see "compactness" defined either for a subset of a topological space, or for a topological space itself. The two definitions are equivalent (in that a subset of a topological space is compact as a subset if and only if it is compact as a space in its own right with the subset topology), but the details are subtly different and should not be mixed up.

If $X$ is just a subset, then there's no reason to except that the finite union equals $\bigcup_I U_i$, becuase $\bigcup_I U_i$ can contain points outside $X$ that are not in the finite union.

On the other hand, if $X$ is the entire topological space, then every $U_i$, and therefore also their union, is by definition a subset of $X$. In that case, claiming that $X\subseteq \bigcup_{\rm finite}U_i$ is the same as claiming that $X=\bigcup_{\rm finite}U_i$, and that is how the definition would usually be phrased. Since we then, obviously, have $$ X = \bigcup_{\rm finite}U_i \subseteq \bigcup_I U_i \subseteq X $$ this forces the finite and infinite union to equal each other.

You then ask,

Why does this happen?

The best answer I can think of to that is: It doesn't necessarily. But when it does happen for all open covers -- and only then! -- we call the space compact. That's what "compact" means.


I guess you have some misunderstanding, the notation $\displaystyle\bigcup_{i=1}^{n}U_{i}$ in this context means, there exists some finite subset $F\subseteq I$ for which $X=\displaystyle\bigcup_{i\in F}U_{i}$. In general, $I$ is not a finite set (if it were, then $X$ is trivially compact). So now the question is, given $X=\displaystyle\bigcup_{i\in I}U_{i}$, whether we can find any finite set $F\subseteq I$ such that $X=\displaystyle\bigcup_{i\in F}U_{i}$.

For instance, let $I$ be the set of all positive integers, and $U_{i}=B_{i}(0)$, $i=1,2,...$, then ${\bf{R}}=\displaystyle\bigcup_{i\in I}B_{i}(0)$, but can you find a finite subset $F$ of $I$ such that ${\bf{R}}=\displaystyle\bigcup_{i\in F}B_{i}(0)$?

  • $\begingroup$ $I$ is not finite, and $X=\displaystyle\bigcup_{i\in I}U_{i}$ and then you find this $X=\displaystyle\bigcup_{i\in F}U_{i}$ (if $X$ is compact), so we can notice that $ \displaystyle\bigcup_{i\in F}U_{i}=X=\displaystyle\bigcup_{i\in I}U_{i}$ $\endgroup$ – user486983 Dec 17 '17 at 2:45
  • $\begingroup$ Yes, to prove that $X$ is compact, you need to find such a finite set. And for the case ${\bf{R}}$, you will fail to find that, because $F$ has the greatest positive integer, say, $N$, if it were going through, then ${\bf{R}}=B_{N}(0)$, this is clearly absurd. $\endgroup$ – user284331 Dec 17 '17 at 2:47
  • $\begingroup$ So not every space $X$ is compact, ${\bf{R}}$ with the usual topology shows this along with the covering $\{B_{i}(0): i=1,2,...\}$ and it fails to find any finite subcover. $\endgroup$ – user284331 Dec 17 '17 at 2:52

The OP should put down whatever book/notes they are reading and try a fresh approach - read this relevant wikipedia definition for compactness.

It starts off by defining compactness for any topological space $X$, and you will see equal signs, not inclusion signs. It then continues with the definition for a subset of a topological space, and you will see inclusion signs, not equal signs.

Well, to say a subset $K$ of a topological space $X$ is compact means that the (induced) topological subspace $K$ is compact. So, of course, if you don't like seeing the inclusion symbols, you can write out the 'induced formulation' using only the equal sign (provided you understand what the induced topology is).

The wikipedia article also states

Compactness is a "topological" property. That is, if ${\displaystyle X\subset Z\subset Y}$, with subset $Z$ equipped with the subspace topology, then $X$ is compact in $Z$ if and only if $X$ is compact in $Y$.

In particular, the induced topology on $X$ is compact if and only if $X$ is a compact subset of $Y$.

Of course if $X$ is a compact space, the subset $X \subset X$ with the induced topology is compact.

Writing all this out makes me want to google
'tautologically treating tortuous topological type things'.


No. $K = [0,1]$ is compact.
$C = \{ (-1,n) : n \in \mathbb N \}$ covers $K$.
$Cf = \{ (-1,2) \}$ is a finite subcover.
The union of $C \neq$ the union of $Cf$.


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