Is a Baire Space necessarily complete?

Question

A complete metric space a Baire space. But is a Baire space necessarily complete?

Answer 1

Hint 1: Prove that an open subspace of a Baire space is a Baire space. Think of an open subspace of $\mathbb R$ that is not complete.

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Hint 2: If $Y$ is a $G_\delta$ subset of $X$, and $X$ is compact Hausdorff or complete metric, then $Y$ is a Baire space in the subspace topology. This shows that the irrationals are a Baire space.

Answer 2

The other examples (like $(0,1)$ or the irrationals) are still completely metrizable (i.e. there is an equivalent metric that is complete). In fact, a subset of a complete metric space is completely metrizable iff it is a $G_\delta$ in that space, and all completely metrizable spaces are Baire.

There are more Baire spaces than just those $G_\delta$'s: any space $X$ that has an open and dense subspace that is Baire is itself Baire, e.g., so the set $\mathbb{R} \times (0,+\infty) \cup \mathbb{Q} \times \{0\}$ in the plane is Baire, but not completely metrizable, as $\mathbb{Q}$ embeds as a closed set into it.

Answer 3

Being "Baire" is really a topological property, so I think that it is not an appropriate question to ask whether Baire spaces are complete as completeness is not a topological property.

Consider for example, $\mathbb{Q}$ with the standard metric topology. This is not a complete space. Nor is it a Baire space. But one can ask the question of whether it is possible to define a metric on $\mathbb{Q}$ such that this new metric induces the same topology as before but this new metric is complete. The answer is no because $\mathbb{Q}$, ignoring its metric and just looking at its topology, is not a Baire space and so if there was a complete metric it would contradict Baire's category theorem.

Another example, consider $X=(-\frac{\pi}{2},\frac{\pi}{2})$. This is a Baire space (because it is locally compact and Hausdorff). We can use the standard Euclidean metric to induce its topology and with that metric it is not complete. But we can also define a new metric $d(x,y) = |\tan x - \tan y|$. Then $(X,d)$ is isometric to $(\mathbb{R},|\cdot |)$ via the map $x\mapsto \tan x$ so with this new metric $X$ is now complete. So the same exact topological space that is Baire is either complete or not complete by changing metrics.