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    $G_\delta$ subspaces of compact Hausdorff or complete metric spaces are Baire

    E-CKXKK

    Area
    Topology
    Topics
    Baire Spaces
    Status
    draft

    Show that if \(Y\) is a \(G_\delta\) set in \(X\), and if \(X\) is compact Hausdorff or complete metric, then \(Y\) is a Baire space in the subspace topology. [Hint: Suppose that \(Y = \bigcap W_n\), where \(W_n\) is open in \(X\), and that \(B_n\) is closed in \(Y\) and has empty interior in \(Y\). Given \(U_0\) open in \(X\) with \(U_0 \cap Y \neq \varnothing\), find a sequence of open sets \(U_n\) of \(X\) with \(U_n \cap Y\) nonempty, such that

    \[ \begin{array}{c} \overline{U}_n \subset U_{n-1}, \\ \overline{U}_n \cap \overline{B}_n = \varnothing, \\ \operatorname{diam} U_n < 1/n \quad \text{in the metric case}, \\ \overline{U}_n \subset W_n.] \end{array} \]

    Solution

    Write \(Y=\bigcap_{n\ge1}W_n\) with \(W_n\) open in \(X\). To prove \(Y\) Baire, let \(B_n\) be closed nowhere dense subsets of \(Y\) and let \(U_0\) be open in \(X\) with \(U_0\cap Y\ne\varnothing\). We construct nonempty open \(U_n\) meeting \(Y\) such that \[ \overline U_n\subset U_{n-1}\cap W_n,\qquad \overline U_n\cap\overline{B_n}^{\,X}=\varnothing, \] and in the metric case also \(\operatorname{diam}U_n<1/n\).

    Indeed, since \(B_n\) has empty interior in \(Y\), the nonempty relatively open set \(U_{n-1}\cap W_n\cap Y\) contains a point outside \(B_n\). Because \(B_n\) is closed in \(Y\), choose an \(X\)-open neighborhood \(V\) of that point whose closure is contained in \(U_{n-1}\cap W_n\) and avoids \(\overline{B_n}^{\,X}\); in the metric case shrink its diameter below \(1/n\). (In the compact-Hausdorff case use regularity.) Set \(U_n=V\).

    If \(X\) is compact Hausdorff, the nested compact sets \(\overline U_n\) have nonempty intersection. If \(X\) is complete metric, choose \(x_n\in\overline U_n\); nestedness and the diameter bound make \((x_n)\) Cauchy, with limit in every \(\overline U_n\). Thus choose \(x\in\bigcap_n\overline U_n\). Since \(\overline U_n\subset W_n\), \(x\in Y\); since \(\overline U_n\) avoids \(B_n\), \(x\notin\bigcup B_n\); and \(x\in\overline U_1\subset U_0\). Hence no nonempty relatively open subset of \(Y\) is covered by countably many nowhere dense closed sets, so \(Y\) is Baire.