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I am baffled with what I am seeing. First, here's what is noted as a definition in my notes

Let $X$ be a set and $A \subseteq X$. A cover of $A$ by subsets of $X$ is a family $(W_i)_{i \in I}$ of subsets of $X$ such that $A \subseteq \cup_{i \in I} W_i$. For $J \subseteq I$ we say that $(W_j)_{j \in J}$ is a subcover of $(W_i)$ if itself is a cover of $A$ by subsets of $X$.

Sure. Got it. However, the following which to me, seems like a definition is given as a lemma

$A \subseteq X$ is compact if and only if every cover of $A$ by open subsets of $X$ has a finite subcover.

That's essentialy the definition of "compactness" .... The proof (forward) I looked at and it still is strange

Suppose $A$ is compact. Let $(W_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. For each $i \in I$, put $U_i = A \cap W_i$. BY the suspace topology, $U_i$ is open in $A$. Then,

$$A=(\cup W_i) \cap A=\cup(W_i \cap A)=\cup U_i$$

Hence $(U_i)$ is an open cover of $A$. Since $A$ is compact, we can choose finite $J \subseteq I$ such that $A=\cup U_j$. But $U_j \subseteq W_j$ for all $j$ so $A \subseteq \cup W_j$

Well, why go through the trouble? I mean it's said that $A$ is compact by assumption i.e. every open cover of $A$ has a finite subcover so if there is an open cover $(W_i)$, then there is a finite subcover for sure.

I mean what is going on with this bizarre proof of a definition? A compactness of a subspace is a lemma? Not given by a definition? I am so confused, then why is a the general definition of a compact space $X$ given without proof? i.e. Why is it not

$X$ is compact if and only if every open cover has a finite subcover and now here's the proof...

As much as the above doesn't make sense since that's how we acknowledge the notion, the proof for $A$ doesn't make sense to me.

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  • $\begingroup$ You can define compactness using the finite intersection property..could he/she have done this? $\endgroup$
    – B. Pasternak
    May 9, 2016 at 16:45
  • $\begingroup$ They might define compact differently. What class is this in? $\endgroup$
    – user223391
    May 9, 2016 at 16:47
  • $\begingroup$ It's in a course called general topology. It's an undergraduate course which is supposedly an introduction to topology... I've touched once on the finite intersection property but not too familiar with it. No have i associted it with compactness. $\endgroup$
    – Kydo
    May 9, 2016 at 16:50
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    $\begingroup$ Please avoid unflattering descriptions in titles when, until proven otherwise, a likely cause of your puzzlement is that YOU misunderstood something. $\endgroup$
    – Did
    May 9, 2016 at 16:55
  • $\begingroup$ you seem to have been irritated with the use of "bizarre" i guess, well thanks for fixing it anyway $\endgroup$
    – Kydo
    May 9, 2016 at 17:10

1 Answer 1

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Your confusion here is that there are two different topologies in play. We have the topology on $X$ and the subspace topology on $A$. The lemma says that $A$ is compact with respect to its subspace topology if and only if "$A$ is compact with respect to the whole topology".

I put this last part in quotes because this second idea of "compactness" isn't usually defined as compactness, just understood to be equivalent to the usual "compactness"

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  • $\begingroup$ So the issue is that it needs to be fitted into the context of subspace topology given the definition in the larger space $X$...? And the proof gives a way of doing that? $\endgroup$
    – Kydo
    May 9, 2016 at 17:11
  • $\begingroup$ If I understand what you are saying, then yes. The problem is that "compact" is a property of a space, not a subset. We tend to use "compact" to describe subsets that are compact when considered as subspaces with the subspace topology. This lemma essentially justifies our loose use of the word. $\endgroup$
    – Sean English
    May 9, 2016 at 17:14

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