Prove that every set and subset with the cofinite topology is compact

Prove that every set with the cofinite topology is compact as well as every subset

Solution. Let $$X$$ be a nonempty set with the cofinite topology and let $$\mathscr{U}$$ be an open cover of $$X$$. Let $$U \in \mathscr{U}$$. Then $$X\setminus U$$ is finite. For every $$a \in X\setminus U$$ let $$U_a$$ be an element of $$\mathscr{U}$$ that contains $$a$$. Then $$\{U\}\cup\{U_a : a ∈ X\setminus U\}$$ is a finite subcover of $$\mathscr{U}$$.

Now I missing the part for the subsets $$E\subseteq X$$. I don't think this refers to the relative topology, but just to any subset of $$X$$ How do I go about it?

• It of course refers to the rel topology and the proof is the same. – JCAA Jul 7 '20 at 22:10
• I thought the statement meant that any subset of X was compact with the original topology, not with the relative one – J.C.VegaO Jul 7 '20 at 22:13
• Well what does it mean to be compact with the original topology? – Severin Schraven Jul 7 '20 at 22:21
• @Severin Schraven That you can extract a finite subcover from a open cover made of open sets of the original topology, ( not made of the intersection of them with the set, like in the relative topology.) The reason I make a difference is because set that is not open in the original topology may be open in the relative. like for example in [-1,1] with the usual topology $\tau$ as a subset of $\mathbb{R}$ . $E=[-1,1 ]$ is not open in $(\mathbb{R},\tau)$ but it is in $(\mathbb{R},\tau_E)$ – J.C.VegaO Jul 7 '20 at 22:27

In the case of compactness it makes no difference whether you use the topology of $$X$$ or the relative topology on the subset.

Proposition. Let $$\langle X,\tau\rangle$$ be any space, let $$K\subseteq X$$, and let $$\tau_K$$ be the relative topology on $$K$$; then $$K$$ is compact with respect to $$\tau$$ iff it is compact with respect to $$\tau_K$$.

Proof. Suppose first that $$K$$ is compact with respect to $$\tau$$, and let $$\mathscr{U}\subseteq\tau'$$ be a $$\tau'$$-open cover of $$K$$. For each $$U\in\mathscr{U}$$ there is a $$V_U\in\tau$$ such that $$U=K\cap V_U$$. Let $$\mathscr{V}=\{V_U:U\in\mathscr{U}\}$$; clearly $$\mathscr{V}$$ is a $$\tau$$-open cover of $$K$$, so it has a finite subcover $$\{V_{U_1},\ldots,V_{U_n}\}$$. Let $$\mathscr{F}=\{U_1,\ldots,U_n\}$$; $$\mathscr{F}$$ is a finite subset of $$\mathscr{U}$$, and

$$\bigcup\mathscr{F}=\bigcup_{k=1}^nU_k=\bigcup_{k=1}^n(K\cap V_{U_k})=K\cap\bigcup_{k=1}^nU_k=K\;,$$

so $$\mathscr{F}$$ covers $$K$$. Thus, $$K$$ is compact with respect to $$\tau'$$.

Now suppose that $$K$$ is compact with respect to $$\tau'$$, and let $$\mathscr{U}\subseteq\tau$$ be a $$\tau$$-open cover of $$K$$. For each $$U\in\mathscr{U}$$ let $$V_U=K\cap U$$, and let $$\mathscr{V}=\{V_U:U\in\mathscr{U}\}$$. $$\mathscr{V}$$ is a $$\tau'$$-open cover of $$K$$, so it has a finite subcover $$\{V_{U_1},\ldots,V_{U_n}\}$$. Let $$\mathscr{F}=\{U_1,\ldots,U_n\}$$; $$\mathscr{F}$$ is a finite subset of $$\mathscr{U}$$, and

$$\bigcup\mathscr{F}=\bigcup_{k=1}^nU_k\supseteq\bigcup_{k=1}^n(K\cap U_k)=\bigcup_{k=1}^nV_{U_k}=K\;,$$

so $$\mathscr{F}$$ covers $$K$$. Thus, $$K$$ is compact with respect to $$\tau$$. $$\dashv$$

• Initially I thought that they were saying that any subset of $X$ was compact, like for example if $X=\mathbb{R}$ with the cofinite topology, then any subset like $\{1\}$ $[1,2] ,(1,2),(1,2], (1,+\infty)$ should be compact, that is not true, is it? – J.C.VegaO Jul 7 '20 at 23:43
• @J.C.VegaO: It is true. You essentially proved it in your question. – Brian M. Scott Jul 7 '20 at 23:46