Show that if \(X=\mathbb R\) is the set of real numbers with the infinite ray topology \(\mathcal T_X=\{(a,\infty)\mid a\in\mathbb R\}\cup\{\varnothing,\mathbb R\}\), then there are compact subsets \(C,D\) of \(X\) such that \(C\cap D\) is not compact.
Solution
Let \[ S=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}, \qquad C=S\cup\{0\}, \qquad D=S\cup\{-1\}. \]
1. Every nonempty subset \(K\subseteq X\) that has a minimum element is compact.
Let \(m=\min K\), and let \(\mathcal U\) be an open cover of \(K\). Choose \(U\in\mathcal U\) with \[ m\in U. \] The nonempty open set \(U\) is either \(X\) or a ray \[ (a,\infty) \] for some \(a\in\mathbb R\).
If \(U=X\), then \(U\) already covers \(K\). Otherwise \(m\in(a,\infty)\), so \(a<m\). Every \(x\in K\) satisfies \(x\ge m>a\), hence \[ x\in(a,\infty)=U. \] Thus in either case the single member \(U\) covers \(K\). Every open cover therefore has a finite subcover, so \(K\) is compact.
2. The sets \(C\) and \(D\) are compact.
Every element of \(S\) is positive, so \[ \min C=0 \qquad\text{and}\qquad \min D=-1. \] Apply 1 to each set.
3. We have\[ C\cap D=S. \]
Both sets contain \(S\). The only point of \(C\) outside \(S\) is \(0\), which is not in \(D\), and the only point of \(D\) outside \(S\) is \(-1\), which is not in \(C\). Hence their intersection is exactly \(S\).
4. The family\[ \mathcal V = \left\{ \left(\frac1n,\infty\right):n\ge2 \right\} \]is an open cover of \(S\).
Each member is open by the definition of the infinite-ray topology. Take any \(1/k\in S\). Choose an integer \[ n>\max\{k,1\}. \] Then \[ \frac1n<\frac1k, \] so \[ \frac1k\in\left(\frac1n,\infty\right). \] Thus every point of \(S\) is covered.
5. The cover \(\mathcal V\) has no finite subcover.
Take finitely many members, with indices \[ n_1,\ldots,n_r, \] and put \[ N=\max\{n_1,\ldots,n_r\}. \] Since the rays are nested, \[ \bigcup_{j=1}^r \left(\frac1{n_j},\infty\right) = \left(\frac1N,\infty\right). \] But \[ \frac1{N+1}\in S \qquad\text{and}\qquad \frac1{N+1}\le\frac1N, \] so this point is not in the finite union. Hence no finite subfamily covers \(S\).
6. Therefore \(C\) and \(D\) are compact while \(C\cap D\) is not compact.
2. By 3, their intersection is \(S\), and <1>4–<1>5 exhibit an open cover of \(S\) with no finite subcover.Thus \(C\cap D\) is not compact, as required.