Check the details of Examples 1 and 2 of §40.
Solution
We verify the assertions in Examples 1 and 2.
Example 1. Every open set \(U\subset X\) is a \(G_\delta\) set, since \[ U=\bigcap_{n=1}^{\infty}U. \]
Now suppose \(X\) is first countable and Hausdorff, and fix \(x\in X\). Let \[ B_1,B_2,\ldots \] be a countable neighborhood basis at \(x\). Replacing \(B_n\) by \[ C_n=B_1\cap\cdots\cap B_n \] we may assume \(C_{n+1}\subset C_n\). Then \[ \{x\}=\bigcap_{n=1}^{\infty}C_n. \] Indeed, \(x\) belongs to every \(C_n\). If \(y\ne x\), Hausdorffness gives a neighborhood \(U\) of \(x\) missing \(y\); some basis element \(B_n\subset U\), hence \(y\notin C_n\). Thus every singleton is \(G_\delta\).
Finally, the top point \(\Omega\) of \(\overline S_\Omega=[0,\Omega]\) is not a \(G_\delta\) set. If \[ \{\Omega\}=\bigcap_{n=1}^{\infty}U_n \] with each \(U_n\) open and containing \(\Omega\), then for each \(n\) there is \(\alpha_n<\Omega\) such that \[ (\alpha_n,\Omega]\subset U_n. \] Because a countable supremum of countable ordinals is countable, \[ \alpha=\sup_n\alpha_n<\Omega. \] Any \(\beta\) with \(\alpha<\beta<\Omega\) then belongs to every \(U_n\), contradicting that their intersection is \(\{\Omega\}\).
Example 2. Let \(A\) be closed in a metric space \((X,d)\), and write \[ U(A,\varepsilon)=\{x:d(x,A)<\varepsilon\}. \] Each \(U(A,1/n)\) is open. Clearly \(A\subset U(A,1/n)\) for every \(n\). Conversely, if \(x\notin A\), then \(A\) is closed, so \(d(x,A)>0\): otherwise \(d(x,A)=0\) would imply \(x\in\overline A=A\). Choose \(n\) with \(1/n<d(x,A)\). Then \(x\notin U(A,1/n)\). Hence \[ A=\bigcap_{n=1}^{\infty}U(A,1/n), \] so every closed subset of a metric space is \(G_\delta\).