# Any uncountable not-compact topological space has uncountable number of compact and noncompact subsets

Question: Prove that every uncountable topological space which is not compact has an uncountable number of subsets which are compact and an uncountable number which are not compact.

Since any finite set of a topological space is compact, $\{x_i\}$ is compact for each $x_i$ in $X$. Therefore there's an uncountable number of compact sets since there are an uncountable number of singleton sets in an uncountable set.

I'm not sure how I would prove that there are an uncountable number of non-compact sets.

• @bof I mistyped the question and edited it for clarity. I'm assuming that the space $(X,T)$ is not compact and is uncountable. – Oliver G Jun 5 '16 at 1:31

The crucial sub-problem is the following. Suppose $A, B$ are both compact. Then what can you say about $A\cup B$?

Given that the whole space $X$ is not compact (as per your recent edit), what does this tell you about cosingletons $X\setminus\{x\}$ ($x\in X$)?

Notice that this argument only works for finite unions: do you see why an infinite (even countable) union of compact sets need not be compact?

In fact, any non-compact space has at least $2^{\aleph_0}$ non-compact subsets.

Let $X$ be a non-compact space; $X$ is infinite, since every finite space is compact. Let $\mathcal K$ be the collection of all compact subsets of $X,$ and let $\mathcal N=\mathcal P(X)\setminus\mathcal K$ be the collection of all non-compact subsets of $X.$ Since the union of two compact sets is compact, the complement of an element of $\mathcal K$ is an element of $\mathcal N;$ thus we get an injection $f:\mathcal K\to\mathcal N$ by setting $f(K)=X\setminus K.$ This shows that $|\mathcal K|\le|\mathcal N|.$ Since $|\mathcal K|+|\mathcal N|=|\mathcal K\cup\mathcal N|=|\mathcal P(X)|=2^{|X|}$ is an infinite cardinal, it follows that $|\mathcal K|+|\mathcal N|=|\mathcal N|$ and so $|\mathcal N|=2^{|X|}\ge2^{\aleph_0}.$

If we assume that $X$ is an infinite Hausdorff space, then we can drop the assumption that $X$ is not compact: any infinite Hausdorff space $X$ has $2^{|X|}$ non-compact subsets. If $X$ is discrete, then every infinite subset of $X$ is non-compact, and $X$ has $2^{|X|}$ infinite subsets. If $X$ is not discrete, then some one-point set $\{x\}$ is not open. Since compact subsets of a Hausdorff space are closed, the set $Y=X\setminus\{x\}$ is not compact, whence it has $2^{|Y|}=2^{|X|}$ non-compact subsets.

• Why is it that in a non-compact set, the complement of a compact subset is a non-compact subset? – Oliver G Jun 5 '16 at 12:34
• @OliverG You are asking why the union of two compact sets is compact? If $\mathcal U$ is an open cover of $A\cup B,$ then it's an open cover of $A$ and of $B,$ and the union of a finite subcover of $A$ and a finite subcover of $B$ is a finite subcover of $A\cup B.$ – bof Jun 5 '16 at 19:49
• I'm asking why the complement of a compact subset is a not compact subset. You wrote, "Since the union of two compact sets is compact, the complement of an element of $K$ is an element of $N$." So the complement of a compact set is a not compact set? Why is this? – Oliver G Jun 6 '16 at 1:00
• @OliverG If the complement of a compact set were a compact set, then the union of the set and its complement (which is the whole space) would be the union of two compact sets, so it would be compact. – bof Jun 6 '16 at 2:14
• So if a space is not compact, then a compact subset has a non-compact complement. So if a space is compact, is this still true? Let $(X,T)$ be compact, Let $A \subseteq X$ is compact, Suppose $X \backslash A$ is compact, Then $X \backslash A \cup A = X$ is compact, which is true. But $[0,10] \backslash [2,6] = [0,2) \cup (6,10]$ which is not compact. – Oliver G Jun 6 '16 at 12:29