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?



Completely Metrizable Space and Baire Theorem
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.


