1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ It of course refers to the rel topology and the proof is the same. $\endgroup$ – JCAA Jul 7 '20 at 22:10
  • $\begingroup$ I thought the statement meant that any subset of X was compact with the original topology, not with the relative one $\endgroup$ – J.C.VegaO Jul 7 '20 at 22:13
  • $\begingroup$ Well what does it mean to be compact with the original topology? $\endgroup$ – Severin Schraven Jul 7 '20 at 22:21
  • $\begingroup$ @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 $(\mathbb{R},\tau_E)$ $\endgroup$ – J.C.VegaO Jul 7 '20 at 22:27
0
$\begingroup$

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 $\tau$. $\dashv$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ – J.C.VegaO Jul 7 '20 at 23:43
  • $\begingroup$ @J.C.VegaO: It is true. You essentially proved it in your question. $\endgroup$ – Brian M. Scott Jul 7 '20 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.