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Let $X$ be a completely metrizable space and let $A\subset X$ be a co-meager subset.

Since $X$ is Baire, we know that there is some dense $G_\delta$ set contained in $A$, say $G\subset A$. But since $X$ is complete and $G$ is $G_\delta$, then $G$ is complete. Hence, as a complete subspace of another complete subspace, $G$ is closed.

But since $G$ is dense, $G=X$ which implies that $A=X$. So we conclude that any co-meager subset of a completely metrizable space, has to be the space itself. This is not true but I can't see where I made a mistake.

Can somebody help me? Thanks

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Your mistake is "since $X$ is complete and $G$ is $G_\delta$, then $G$ is complete, hence . . . $G$ is closed." Nope. $\mathbb R$ is a complete metric space, and $(0,1)$ being an open set is a $G_\delta,$ but $(0,1)$ is not closed in $\mathbb R.$ You see, $(0,1)$ is completely metrizABLE (being homeomorphic to $\mathbb R$), but it is not complete with respect to the usual metric of $\mathbb R.$ A $G_\delta$ subset $G$ of a complete metric space $X$ is completely metrizable, there is a complete metric on $G$ which induces the topology which $G$ inherits as a subspace of $X,$ but in general it's not the same as the metric in $X.$ It can't be, unless $G$ happens to be closed in $X.$

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  • $\begingroup$ Thanks to you it is now very clear the silly (but for me, subtle) mistake I made in my reasoning. $\endgroup$ – valls Sep 25 '16 at 8:17

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