Show that in the finite complement topology on \(\mathbb{R}\), every subspace is compact.
If \(\mathbb{R}\) has the topology consisting of all sets \(A\) such that \(\mathbb{R} - A\) is either countable or all of \(\mathbb{R}\), is \([0, 1]\) a compact subspace?
Solution
Let \(Y\subset \mathbb R\), with the subspace topology inherited from the finite-complement topology on \(\mathbb R\). If \(Y\) is finite it is compact. Suppose \(Y\) is infinite. Then a nonempty open set in \(Y\) has the form \[ Y\cap U=Y\setminus F \] with \(F\) finite. Thus the subspace topology on \(Y\) is again the finite-complement topology. Given an open cover \(\{U_i\}_{i\in I}\) of \(Y\), choose one nonempty member \(U_{i_0}=Y\setminus F\). The finite set \(F\) is covered by finitely many further members of the cover, so together with \(U_{i_0}\) these form a finite subcover. Hence every subspace is compact.
No. In the countable-complement topology, the subspace topology on \([0,1]\) is the countable-complement topology on \([0,1]\). Choose distinct points \(x_n\in[0,1]\), and set \[ U_n=[0,1]\setminus\{x_m:m\ge n\}. \] Each \(U_n\) is open in the subspace, the family is increasing, and \(\bigcup_nU_n=[0,1]\). But a finite subfamily has union equal to its largest member \(U_N\), which omits infinitely many points \(x_m\) for \(m\ge N\). Thus there is no finite subcover.