# Counterexample to the Converse of Baire's Category Theorem

In a lecture on Baire's Category Theorem at Indian Institute of Tech, it was mentioned that the converse of Baire's: Every non meagre (second category) space is complete, is not true, and that a proof of the existence of an incomplete non meagre (second category) space was given by Bourbaki.

(N.B. The course used a weaker form of Baire's than usual: Every complete space is non meagre (of second category).)

Question: First of all I have failed to find the proof mentioned, but would also like to ask if anyone can and would give me a short version of why this is, and also tell me what's wrong (the existence of a known proof suggests there is) with the following trivial counterexample:

The interval $$(0,1)$$ is incomplete as a metric subspace of $$\mathbb{R}$$, yet it is non meagre.

Any open subset of a complete metric space (more generally and $$G_{\delta}$$ subset) has an equivalent metric which makes it complete. So it is non-meagre.
In the case of $$(0,1)$$ such a metrc is deined by $$D(x,y)=|x-y|+|\frac 1 {d(x)} -\frac 1{d(y)}|$$ where $$d(x) =\min \{{x, 1-x}\}$$.
• I'm not sure I entirely understand, this shows that $(0,1)$ is non-meagre in $((0,1),D)$, by completeness of $((0,1),D)$, fine. But, in the converse of Baire's, I thought it was enough to exhibit an example of an incomplete metric space which is non-meagre: My example was $(0,1)\subset \mathbb{R}$ with the usual metric. Either $(0,1)$ is meagre in it self, or it is complete, otherwise what is wrong with the example? Would the converse somehow be more explicitly: Every second cat. space is isomorphic to a complete space? And thus my example would not suffice? – Christopher.L Feb 12 at 13:04
• Also, I guess I thought since 'complete' would be a metric property, that isometries would preserve completeness, but I found out that equivalent metrics does not imply isometric spaces. However, we do still say that $(0,1)\subset \mathbb{R}$ is incomplete, so the question remains whether I have misunderstood the converse of the theorem or non meagre. – Christopher.L Feb 12 at 13:14