The version of the Baire Category Theorem I have in mind is the statement that a countable intersection of dense open subsets of a complete metric space is dense. The question is: is it likewise true that the countable intersection of dense open subsets of a complete uniform space is dense?

A look at the proof suggests that the answer should be no---it makes crucial use of the fact that the canonical uniform structure on a metric space is generated by the $\varepsilon$-balls of radius $\frac{1}{m}$ for $m\ \in \mathbb{Z}^+$. I thus started looking at complete topological groups that were not first-countable for a counter-example, with no luck (topological groups because they provide easy examples of uniform spaces, and not first-countable because first-countable topological groups have uniformities generated by a countable collection of covers (and also because (I think?) first-countable topological groups are metrizable).)

Surprisingly, a quick Google search only pulled up one source that seems to mention this at all---Joshi in his Introduction to General Topology states (pg. 363)

Unfortunately, there seems to be no analogue of the Baire category theorem for complete uniform spaces.

but yet does not provide a counter-example either.

Also, I doubt this will make too much of a difference, but I would prefer the counter-example to be $T_0$. Finding an example of a highly pathological space that is, uhh, pathological, isn't exactly what I'm looking for.

  • $\begingroup$ Minor thing: First-countable $T_0$ topological groups are indeed metrisable. For topological groups, having a countable basis at the identity is equivalent to pseudo-metrisability. $\endgroup$ Jul 31, 2015 at 16:25

2 Answers 2


Functional analysis abounds with such examples. There are many complete locally convex spaces which are countable unions of closed subspaces with empty interior. One such is the space of smooth functions on the line which have compact support, the test functions of L. Schwartz. An even simpler one is the space of finite sequences with the locally convex inductive limit topology as the union of finite dimensional spaces.

Added as an edit: The completeness of the first space follows from the fact that it is a strict $LF$-space and these are complete by a result of Dieudonné and Schwartz, of the second from the fact that it is the strong dual of the (nuclear) Fréchet space of all sequences.

  • $\begingroup$ Wow. Thank you. Thinking about it 'dually' (in terms of closed subsets instead of open subsets) makes this incredibly obvious. Now I feel silly =P $\endgroup$ Jul 25, 2015 at 18:12
  • $\begingroup$ Actually, might I ask what your proof of completeness is? It is well-known that this test function space is sequentially-complete, but I am having difficulty in proving that every cauchy net converges. Moreover, I now recall Vobo stating on an old question of mine that it is in fact not cauchy complete (see math.stackexchange.com/questions/218709/…). If this does indeed turn out to fail, do you have other examples that will work? $\endgroup$ Jul 26, 2015 at 19:41
  • $\begingroup$ First of all, it IS complete. This was proved by Dieudonné and Schwartz in their seminal paper and is by no means trivial. However, for your purposes the second example would suffice and is much simpler. In fact, a fecund family displaying the kind of phenomenon that you are after is that of the complete $DF$-spaces of Grothendieck which never have the Baire property since they are countable unions of bounded sets---well hardly ever since the usual degenerate examples have to be taken into consideration. $\endgroup$
    – priel
    Jul 26, 2015 at 20:32
  • $\begingroup$ The paper referred to above is MR 0038533. $\endgroup$
    – priel
    Jul 26, 2015 at 20:41
  • 1
    $\begingroup$ I have added some hints on why my examples are complete---you can't expect me to write down detailed proofs of results which are available in the literature on this site. Even if I wanted to, it would take me forever since I am working from my iPad. $\endgroup$
    – priel
    Jul 27, 2015 at 5:41

It is very common for a topological space to be a complete uniform space in some uniformity, but it is less common for a topological space to satisfy the Baire category theorem since the proof of the Baire category theorem makes an essential use of countability of the metric rather than the uniformity..

Every paracompact space has a compatible complete uniformity simply by taking the uniform covers to be open covers and the entourages to be the neighborhoods of the diagonal. In fact, the paracompact spaces are precisely the spaces that have a compatible uniformity that satisfies a stronger form of completeness. A uniform space is said to be supercomplete if its hyperspace is complete. A topological space has a compatible supercomplete uniformity if and only if it is paracompact.

A space whose cardinality is below the first measurable cardinal is completely uniformizable if and only if the space is realcompact.

Every countable completely regular space without any isolated points $X$ can be given a compatible complete uniformity, but $X$ is never a Baire space. For example, $\mathbb{Q}$ can be given a compatible complete uniformity, but $\mathbb{Q}$ is the countable union of nowhere dense subspaces.

Every good general topology text will tell you that an $F_{\sigma}$-set in a paracompact space is paracompact. The spaces mentioned by Priel in his answer are paracompact since they are countable unions of closed subspaces of a paracompact space (the space of all smooth functions with compact support is metrizable since its topology is induced by countably many seminorms, namely $Sup_{x\in\mathbb{R}} |f^{(n)}(x)|$ and every metrizable space is paracompact).

If $X$ is a completely regular space, then let $Ba(X)$ denote the $\sigma$-algebra of Baire sets. In other words, $Ba(X)$ is the $\sigma$-algebra on $X$ generated by the zero sets (Recall that a zero-set is a set of the form $f^{-1}[\{0\}]$ for some continuous $f:X\rightarrow\mathbb{R}$ and that the complement of a zero set is a cozero set).

$\mathbf{Lemma}$ A completely regular space $X$ is realcompact if and only if each $\sigma$-complete ultrafilter on $Ba(X)$ is of the form $\{R\in Ba(X)|x_{0}\in R\}$ for some

$\mathbf{Proposition}$ Every $F_{\sigma}$-set in a realcompact space is also realcompact.

$\mathbf{Proof}:$ Suppose that $X$ is a realcompact space and $C_{n}\subseteq X$ is a closed subspace for each natural number $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Let $\mathcal{M}\subseteq Ba(D)$ be a $\sigma$-complete ultrafilter. Then define $\mathcal{N}=\{R\in Ba(X)|R\cap D\in\mathcal{M}\}$. Then $\mathcal{N}$ is a $\sigma$-complete ultrafilter on $Ba(X)$. Therefore, $\mathcal{N}=\{R\in\mathcal{M}|x_{0}\in R\}$ for some $x_{0}\in X$. I claim that $x_{0}\in D$.

Suppose to the contrary that $x_{0}\not\in D$. Then since $x_{0}\in\overline{D}$ and $X$ is completely regular, for all $n$ there is some cozero set $U_{n}\subseteq X$ with $x_{0}\in U_{n}$ but $U_{n}\cap C_{n}=\emptyset$. Therefore, $U_{n}\in\mathcal{N}$ for each $n$, so $\bigcap_{n}U_{n}\in\mathcal{N}$ as well. However, since $\bigcap_{n}U_{n}\in\mathcal{N}$, we have $\emptyset=D\cap\bigcap_{n}U_{n}\in\mathcal{M}$ which is a contradiction. Therefore, $x_{0}\in D$.

I now claim that $\mathcal{M}=\{R\in Ba(D)|x_{0}\in R\}$. Suppose that $R\in Ba(D)$ and $x_{0}\in R$. Then there are open subsets $V_{n}\subseteq D$ such that $x_{0}\in V_{n}$ for all $n$ but where $\bigcap_{n} V_{n}\subseteq R$. Therefore, for all $n$, there is some open subset $V_{n}^{\sharp}\subseteq X$ such that $V_{n}=V_{n}^{\sharp}\cap D$. Therefore, since $x_{0}\in V_{n}^{\sharp}$, for all $n$, there is a cozero set $W_{n}^{\sharp}\subseteq X$ such that $x_{0}\in W_{n}^{\sharp}\subseteq V_{n}^{\sharp}$. Let $W_{n}=W_{n}^{\sharp}\cap D$. Then $W_{n}^{\sharp}\in\mathcal{M}$ for all $n$, so $W_{n}=W_{n}^{\sharp}\cap D\in\mathcal{N}$. Therefore $\bigcap_{n\in\omega}W_{n}\in\mathcal{N}$. However, since $W_{n}\subseteq V_{n}$ and $\bigcap_{n\in\omega}W_{n}\subseteq\bigcap_{n}V_{n}\subseteq R$, we have $R\in\mathcal{N}$ as well. We therefore conclude $\mathcal{N}=\{R\in Ba(D)|x_{0}\in R\}$. Therefore $D$ is realcompact. $\mathbf{QED}$

Using the above proposition, one can obtain a realcompact space that does not satisfy the Baire category theorem from any completely regular space that does not satisfy the Baire category theorem: Suppose that $X$ is a completely regular space which is the union of countably many nowhere dense sets $(L_{n})_{n\in\omega}$. Let $Y$ be a realcompact space such that $X$ is a dense subspace of $Y$ (for example, $Y$ could be the Hewitt-realcompactification of $X$). Let $C_{n}=CL_{Y}(L_{n})$ for all $n$. Let $D=\bigcup_{n\in\omega}C_{n}$. Then $D$ is an $F_{\sigma}$-set in $Y$, so $D$ is realcompact. However, each $C_{n}$ is nowhere dense in $D$, so $D$ is not a Baire space. In fact, since every $F_{\sigma}$ subset of a paracompact space is paracompact, if the space $Y$ is paracompact, then the obtained space $D$ would be paracompact but a countable union of nowhere dense sets.

Let me close with a couple propositions that further show that non-Baire uniform spaces and non-Baire complete uniform spaces are quite common.

$\mathbf{Proposition}$ Suppose that $X$ is a regular space without any isolated points such that for each $x\in X$ there is a collection $\mathcal{U}$ of pairwise disjoint open sets and $x_{U}\in U$ for each $U\in\mathcal{U}$ (i.e. every point is the limit of a discrete set) such that $x\in\overline{\{x_{U}|U\in\mathcal{U}\}}\setminus\{x_{U}|U\in\mathcal{U}\}$. Then for each $x_{0}\in X$ there is a non-Baire subspace containing $x_{0}$.

$\mathbf{Proof}$ We shall construct a sequence of disjoint sets $(C_{n})_{n\in\omega}$ along with open neighborhoods $U_{x}$ of $x$ for each $x\in X$ by induction on $n$.

For $n=0$, let $x_{0}\in X$, let $C_{0}=\{x_{0}\}$ and let $U$ be an open neighborhood of $x_{0}$.

Now assume that $n>0$ and $C_{n-1}$ along with $U_{x}$ for $x\in C_{n-1}$ has been constructed already. Then for each $x\in C_{n-1}$, let $L_{x}\subseteq U_{x}$ be a subset and let $(U_{z})_{z\in L_{x}}$ be a collection of sets with $\overline{U_{z_{1}}}\cap\overline{U_{z_{2}}}=\emptyset$ and such that $z\in U_{z}\subseteq\overline{U_{z}}\subseteq U_{x}$ for each $z\in L_{x}$. Let $C_{n}=\bigcup_{x\in C_{n-1}}L_{x}$

Since each $x\in C_{n-1}$ is non-isolated in $C_{n-1}\cup C_{n}$ and $U_{x}\cap C_{n-1}=\{x\}$, the set $C_{n-1}$ is nowhere dense in $C_{n-1}\cup C_{n}$. Therefore if $D=\bigcup_{n\in\omega}C_{n}$, then each set $C_{n}$ is is nowhere dense in $D$, so $D$ is not a Baire space. $\mathbf{QED}$.

$\mathbf{Proposition}$ Suppose that $X$ is a normal space without any isolated points such that for each nowhere dense subspace $C\subseteq X$ there is a collection $\mathcal{U}$ of pairwise disjoint open subsets of $X$ along with closed nowhere dense subsets $C_{U}\subseteq U$ such that $C\subseteq\overline{\bigcup\{C_{U}|U\in\mathcal{U}\}}\setminus\bigcup\{C_{U}|U\in\mathcal{U}\}.$ Then for each nowhere dense subspace $C\subseteq X$ there is a $F_{\sigma}$ set $D\subseteq X$ with $C\subseteq D$ and where $D$ is a non-Baire space and $D$ is a $F_{\sigma}$ subset of $X$.

  • $\begingroup$ Isn't it Stone's theorem that metric spaces are paracompact? Hence how would it be possible that a topological space has a compatible supercomplete uniformity if and only if it is paracompact? $\endgroup$ May 2, 2017 at 15:00
  • $\begingroup$ You are correct in saying that every metric space is paracompact including the non-complete metric spaces. The compatible supercomplete uniformity however on a paracompact space however is not required to be generated by the metric. For example, $\mathbb{Q}$ is paracompact. However, by the Baire category theorem $\mathbb{Q}$, cannot be given a compatible complete metric. However, $\mathbb{Q}$ can be given a compatible supercomplete uniformity which is not induced by any metric on $\mathbb{Q}$. $\endgroup$ May 2, 2017 at 15:16
  • $\begingroup$ Would you mind explaining why $\mathbb{Q}$ has a complete uniformity? The uniformity induced by the usual metric seems to be incomplete, right? $\endgroup$
    – ALife
    Apr 1, 2022 at 15:34
  • $\begingroup$ @ALife. The set $\mathbb{Q}$ does not have a complete metric by the Baire category theorem. The collection of all partitions of $\mathbb{Q}$ into open sets generates a complete (but non-metric) uniformity. If $F$ is a filter that is Cauchy with respect to this uniformity, then $F$ is Cauchy with respect to the Euclidean metric on $\mathbb{Q}$, so $F\rightarrow\alpha$ for some real number $\alpha$. However, one can easily show that $\alpha$ cannot be an irrational number. $\endgroup$ Apr 1, 2022 at 15:49
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    $\begingroup$ Thanks for the additional explanation and the reference. Also realized that your "fine uniformity" is a relatively standard term (presumably the finest uniformity generating the topology, as defined in Willard's General Topology ). Your $d_{x_0}$ also seems to be a non-0-1 pseudometric that achieves similar completeness; that's interesting. Will read over these topics. Thanks! $\endgroup$
    – ALife
    Apr 17, 2022 at 14:03

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