# Completely Metrizable Space and Baire Theorem

Is well know that completely metrizable spaces are Baire's spaces. Reciprocally, if $X$ is a Baire's metric space, then $X$ is completely metrizable?

• In Engelking's "General topology", exercise 4.3C(b) asks (if I interpret the terminology correctly) to give an example of a subset of $\mathbb R^2$ that is Baire but not completely metrizable. Since $\mathbb R^2$ is complete, its subset is completely metrizable if and only if it is $G_\delta$. Engelking gives a hint: use that a space is Baire if it has a Baire dense subset. – Igor Belegradek Dec 1 '13 at 15:16

No, in order for a subspace of a complete metric space to be completely metrizable it is necessary and sufficient for it to be $G_\delta$. There are only $2^{\aleph_0}$ many $G_\delta$ subsets of $\mathbb{R}^2$ but there are $2^{2^{\aleph_0}}$ sets in $$\{X \subseteq \mathbb{R}^2 : (0,1)\times(0,1) \subseteq X \subseteq [0,1]\times[0,1]\},$$ each of which is a metrizable Baire space.
• So a concrete example is $$X:=\big( (0,1)\times(0,1)\big)\cup \big( \{0\}\times([0,1]\cap\mathbb{Q})\big),$$ that is not a $G_{\delta}$ because its trace on $\{0\}\times [0,1]$ is not. – Pietro Majer Dec 1 '13 at 16:33