Examples of closed subspaces of Baire spaces that fail to be Baire?

I am looking for some nice examples of Baire spaces containing closed subspaces that fail to be Baire.

Clearly, $X$ should not satisfy either one of the standard hypotheses for the Baire category theorem: local compactness or complete metrizability or Čech completeness because all these properties are hereditary with respect to closed subspaces.

An interesting example of a Baire metrizable space that fails to be completely metrizable is given in an answer to What are some motivating examples of exotic metrizable spaces. This example contains $\mathbb Q$ as a closed subspace.

It would be nice to see some further examples.

Added: I would prefer to have examples of high regularity (at least Hausdorff, preferably Tychonoff).

Thanks!

2 Answers

If I right understood the notation of the book "Baire spaces" by R.C Haworth and R.A. McCoy then there is a simple example of a Baire space $\mathbb{ R^2\setminus ((R\setminus Q)\times \{0\}})$ with closed non-Baire subset $\mathbb Q\times\{0\}.$

• Thanks a lot, that's a neat example and it's much simpler than expected! Thanks also for the reference, I didn't know about that book. Now I wish someone could come up with an example not involving $\mathbb{Q}$ (or any other dense first category subset of $\mathbb{R}^n$) :-) – Curt F. Jul 26 '13 at 20:06
• It seems that similarly to this example we can embed each Tychonoff space $Y$ as a closed subset into a Baire space $X$. Let $bY$ be an arbitrary compactification of $Y$. Put $X=bY\times [0;1]\setminus (bY\setminus Y)\times\{0\}$. – Alex Ravsky Jul 26 '13 at 20:20
• The link is broken, but you can find a pdf of the book here: pldml.icm.edu.pl/pldml/element/… – Francis Adams Jun 8 at 2:25

Consider the euclidean topology $\tau$ on the rationals and some point $x$ which is not contained in $\mathbb{Q}$. The space $Q := \mathbb{Q} \cup \{x\}$ with the topology $\tau' := \{ O \cup \{x\} : O \in \tau \} \cup \{ \emptyset \}$ is a Baire space since {x} is dense and contained in every dense subset of $Q$ - so any intersection of dense subsets of $Q$ is again dense. Furthermore $\mathbb{Q}$ is a closed subspace of $Q$ which fails to be Baire - so $(Q,\tau')$ provides an example to your question.

• Thank you very much. I forgot to mention explicitly that I'd prefer to have some examples with higher separation properties in order to exclude easy examples such as this one. Nevertheless, I appreciate your help and I hope its okay that I added this requirement to the question. – Curt F. Jul 25 '13 at 19:55
• No problem. I would also like to see more natural examples. :) – Dune Jul 26 '13 at 8:57
• Really nice answer. One moment is not clear to me: why any dense subset of $Q$ contains $x$? – ZFR Nov 8 '19 at 4:14
• @ZFR Since $\mathbb{Q}$ is closed in $Q$, no subset of $\mathbb{Q}$ can be dense in $Q$. – Dune Nov 8 '19 at 9:26