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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.

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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 $d(x) =\min \{{x, 1-x}\}$.

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  • $\begingroup$ 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? $\endgroup$ – Christopher.L Feb 12 at 13:04
  • $\begingroup$ 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. $\endgroup$ – Christopher.L Feb 12 at 13:14
  • $\begingroup$ You have not misunderstood the Converse. It is quite trivial that Converse of BCT is false. I just wanted to add more information to your example. $\endgroup$ – Kavi Rama Murthy Feb 12 at 13:20
  • $\begingroup$ Ok, so my example is a counterexample to conv. of Baire's? And, then perhaps the lecturer gave an over complicated counterexample by referring to Bourbaki, and it was not Bourbaki's intention to merely prove there are incomplete second category spaces, but it was part of something else, and the lecturer merely remembered and gave that reference and did not think of (or disregarded) the more trivial case. $\endgroup$ – Christopher.L Feb 12 at 13:35
  • $\begingroup$ Ok, the reference was apparently (somewhat lost in translation to: 'Was shown by Bourbaki') to an exercise (not a proof) in Bourbaki, which I cannot find since 'Elements of Mathematics' is pretty vast. So, it is possible that the exercise was just that simple after all. $\endgroup$ – Christopher.L Feb 12 at 13:57

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