# Why does having fewer open sets make more sets compact?

In functional analysis and in PDE theory, we are interested in proving existence results. Such results are generally obtained on some compact space for some given topology. And this is the reason why in functional analysis, we are always impoverishing topologies: we begin with the strong topology on, say, $E'$, induced by the norm of $E$; then we have the weak topology $\sigma(E',E'')$ and then we have the star weak topology $\sigma(E',E)$. H. Brézis said in a remark of his book that the less open sets a topology has, the more compact sets it is susceptible to have. Does somebody can help me to understand more about this last statement?

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Note that if $\tau \subset \sigma$ and $K \subset X$ is compact with respect to $\sigma$ then $K$ is compact with respect to $\tau$. I don't think that anything deeper is intended here. –  t.b. Oct 18 '11 at 13:35
To illustrate: the unit ball in $E'$ is compact with respect to the norm if and only if $E$ is finite dimensional. It is compact with respect to the weak topology if and only if $E$ is reflexive and it is always compact with respect to the weak$^{\ast}$-topology. This corresponds to the statement by Brézis, as $\sigma(E',E) \subset \sigma(E',E'') \subset \sigma_{\|\cdot\|_{E'}}$. –  t.b. Oct 18 '11 at 13:42
The less open sets a topological space has, the less possible open covers its every subspace has. Now recall the definition of a compact space: a space is compact if from its every open cover one can choose a finite cover. –  Damian Sobota Oct 18 '11 at 13:43
Remember that a set is compact if every cover has a finite subcover. Intuitively, if you have "fewer" open sets, then a cover would have to consist of "larger" open sets, so you have a "better chance" of being able to find a finite subcover since open sets tend to be "larger", and there's fewer of them. –  Arturo Magidin Oct 18 '11 at 13:44
@everybody: Thank you! –  Benjamin Oct 18 '11 at 13:47

A topological space X is called compact if each of its open covers has a finite subcover. That is, if for every family $\left\{ U_\alpha \right\}_{\alpha \in I}$ of open sets of $X$ such that $X = \bigcup_{\alpha \in I} U_\alpha$, we can find a finite subset $\alpha_1, \dots , \alpha_n \in I$ such that $X = U_{\alpha_1} \cup \dots \cup U_{\alpha_n}$.

So, the difficulty for a space to be compact is the number of possible open covers $\left\{ U_\alpha \right\}_{\alpha \in I}$ it has: if $X$ had just one of these open covers, we would have to find just one open finite subcover of it. For each open cover we add to $X$, we have to solve the problem of finding a finite open subcover again.

But the number of possible open covers of a space depends on the number of its open sets: the more open sets $X$ has, the more open covers we can form with them.

Two extreme examples could be:

1. Take any space $X$ with the trivial topology (just $X$ and $\emptyset$ as open sets). It has just one open cover: $X$ itself. So, it's easy to find a finite subcover: $X$ itself. So $X$ is a compact space.
2. Take any space $X$, with an infinite number of points, with the discrete topology (every subset of $X$ is an open set). We have there a lot of open covers. Particularly, we have the open cover of all the points of $X$, $\left\{ \left\{ x\right\} \right\}_{x\in X}$, which has no finite subcover. So $X$ cannot be compact.

EDIT. See the nice remark t.b. has added, counting the number of compact subsets in both cases.

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In view of the question it's probably worth pointing out that in $X$ with the trivial topology, all subsets are compact, while in $X$ with the discrete topology only finite subsets are compact. So in the trivial topology there are a many more compact subsets than in the discrete one as soon as $X$ is infinite ($2^{|X|}$ compact subsets versus $|X|$). –  t.b. Oct 18 '11 at 13:58