Let \(\mathcal B=\{[a,b):a,b\in\RR\}\); let \(\RR_\ell\) denote \(\RR\) equipped with the topology generated by the basis \(\mathcal B\) (you do not need to prove \(\mathcal B\) is a basis). Decide (with proof) if the following sets are compact in \(\RR_\ell\):
\(A=\{1/n:n\in\NN\}\cup\{0\}\).
\(B=\{-1/n:n\in\NN\}\cup\{0\}\).
Solution
1. (a) \(A\) is compact in \(\RR_\ell\).
Let \(\mathcal U\) be an open cover of \(A\), and choose \(U_0\in\mathcal U\) with \(0\in U_0\). Some basic set \([a,b)\) with \(a\le0<b\) lies in \(U_0\), so \([0,b)\subseteq U_0\). Then \(1/n\in U_0\) for all \(n>1/b\), and finitely many members of \(\mathcal U\) cover the remaining points \(1,\tfrac12,\ldots,\tfrac1N\); with \(U_0\) they form a finite subcover.
2. (b) \(B\) is not compact in \(\RR_\ell\).
The sets \(U_0=[0,1)\) and \(U_n=[-1/n,-1/(n+1))\) for \(n\ge1\) are open in \(\RR_\ell\) and cover \(B\). \(U_0\cap B=\{0\}\) and \(U_n\cap B=\{-1/n\}\) for \(n\ge1\), so a finite subfamily covers only finitely many points of the infinite set \(B\).
3. Q.E.D.
Steps 1 and 2 decide (a) and (b).