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I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power.

Here's the theorem (with proof) and two applications:

(Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.

Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 > 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$.

Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 > 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$

Here is one application (taken from here):

Claim: $[0,1]$ contains uncountably many elements.

Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in (0,1)} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable.

And here is another one (taken from here):

Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.

Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem."

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2… – user38268 Jul 2 '12 at 13:34
In the proof that $[0,1]$ is uncountable you mean to say that it is the union of $[0,1]$ and not $(0,1)$. Otherwise you are missing two points. – Asaf Karagila Jul 2 '12 at 22:27
A nonempty complete metric space $X$ is not a countable union of nowhere dense sets. – nullUser Jul 3 '12 at 1:50
A nice survey paper: MR1640007 (99h:26012). Jones, Sara Hawtrey. Applications of the Baire category theorem. Real Anal. Exchange 23 (2), (1997/98), 363–394. It should be available through Project Euclid. – Andrés E. Caicedo Apr 12 '13 at 19:18

20 Answers 20

If $P$ is an infinitely differentiable function such that for each $x$, there is an $n$ with $P^{(n)}(x)=0$, then $P$ is a polynomial. (Note $n$ depends on $x$.) See the links in the comments.

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Link, taken from comments:… – sdcvvc Jul 4 '12 at 12:03

The uniform boundedness principle of Functional Analysis is a very important application of the Baire Category Theorem.

Added: (t.b.) See also Sokal's A really simple elementary proof of the uniform boundedness theorem for a proof without Baire.

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Thanks : ) So far my favourite answer. (But of course I upvoted all of them.) – Rudy the Reindeer Jul 2 '12 at 13:57
The other one of the big theorems from beginning functional analysis that uses the Baire Category Theorem is the open mapping theorem – Francis Adams Jul 2 '12 at 14:05
And its friend the closed graph theoerem. – ncmathsadist Jul 2 '12 at 14:07
@FrancisAdams Ooh, nice, thank you! Now I have two favourite answers. Why don't you make this comment into an answer so that I can upvote it? – Rudy the Reindeer Jul 2 '12 at 14:08
@ MattN It looks like someone already did. – Francis Adams Jul 2 '12 at 14:23

Let $I=[0,1]$ and $\mathcal{C}(I)= \{ f : I \to \mathbb{R} \ \text{continuous} \}$ with the topology of uniform convergence. Then the set of nowhere differentiable functions over $I$ is dense in $\mathcal{C}(I)$.

The same thing holds in $\mathcal{C}(I)$ for the set of nowhere locally monotonic functions.

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It's in fact possible to prove the slightly stronger result that the set of functions in $\mathcal C(I)$ that are differentiable at a single point is meagre (or "of the first category") in $\mathcal C(I)$ – kahen Aug 9 '12 at 12:10

$\overline{\mathbb Q_p}$ is not complete with respect to the $p$-adic absolute value. This follows from the fact that $\overline{ \mathbb{Q}_p}$ has countably infinite dimension over $\mathbb{Q}_p$ which can be proved using Krasner's lemma.

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There exist $2\pi$-periodic continuous functions whose Fourier series diverge on an uncountable set.

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I'm surprised that you have only one favourite! :-) – Willie Wong Jul 3 '12 at 9:24

It can show that an infinite dimensional Banach space has no countable basis.

Firstly, assume that the Banach space $V$ has countable basis $\{x_1,x_2,\dots\}$, and let $V_n=\operatorname{span}\{x_1,x_2,\dots,x_n\}$. It is not difficult to show that $V_n$ are closed and nowhere dense but by Baire category, $\cup V_n=V$ is impossible. As a result,$V$ must has uncountable basis.

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You mean a countable Hamel basis, and that is exactly what Matt wrote in the question about polynomials. – Asaf Karagila Jul 2 '12 at 15:37
right,they are essentially the same.Thanks for pointing out this. – Ben Jul 2 '12 at 15:41

The rationals are not completely metrizable.

Proof: Since the rationals have no isolated points, $\mathbb Q\setminus\{q\}$ is dense and open for every $q$, but $\bigcap_{q\in\mathbb Q}\mathbb Q\setminus\{q\}$ is an intersection of countably many open dense sets which is empty.

One nice corollary from this (see Nate Eldredge's comment below) is that the rationals are not a $G_\delta$ set of real numbers. Thus we have an example of an $F_\sigma$ which is not $G_\delta$.

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This one? – Rudy the Reindeer Jul 2 '12 at 13:30
What's long about "since it's the union of its countably many points and no point is open"? :) – t.b. Jul 2 '12 at 13:44
@t.b. I am also attending a lecture and was trying to prove something else in my head (and yes, the generic topological space is Hausdorff). Doing that whilst typing an answer is highly nontrivial! – Asaf Karagila Jul 2 '12 at 14:12
@AsafKaragila You could still add the proof... – Rudy the Reindeer Jul 2 '12 at 19:39
You don't need Lavrentyev's theorem here. If $\mathbb{Q}$ were $G_\delta$ in $\mathbb{R}$ it would be a dense $G_\delta$, in particular comeager. But $\mathbb{Q}$ is also meager. This would imply that $\mathbb{R}$ is meager, which by BCT it is not. – Nate Eldredge Jul 3 '12 at 1:05

I'm partial to the Principle of Dependent Choices, myself (which is equivalent to the BCT).

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Drats. I was gonna post that one when I got home, but I see I got here seven minutes too late! Damn buses! :-) – Asaf Karagila Jul 2 '12 at 17:16
Blair's article is unique. Short and concise. This piece of mathematical work is very special for me because was the first article that I read. – Paulo Henrique Jul 3 '12 at 23:44
@Fëanor: I haven't read it. What is the title and where can I find it? – Cameron Buie Jul 4 '12 at 15:16
@CameronBuie the article is: The Baire category theorem implies the principle of dependent choices, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., v. 25 n. 10 (1977), pp. 933–934, by Charles Blair. But it's quite hard to find it. See more reference about it here… – Paulo Henrique Jul 4 '12 at 17:12

The open mapping theorem and closed graph theorem of functional analysis are two vital applications.

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Here is another cool one:

Theorem. There exists a continuous function $f:[0,1] \to \mathbb{R}$ that is not monotone on any interval of positive length.

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Let $(X,d)$ a compact metric space and $V$ a closed subspace of $C(X)$, vector space of continuous functions with real values, endowed with the supremum norm. We assume that each function of $V$ is Hölderian, that is, for all $f\in V$, we can find $C>0$ and $0<\alpha< 1$ such that $$\forall x,y\in X,\quad |f(x)-f(y)|\leqslant C\cdot d(x,y)^{\alpha}.$$ Then $V$ is finite dimensional.

Define $$F_n:=\bigcap_{x,y\in [0,1]}\{f\in V,|f(x)-f(y)|\leqslant n\cdot d(x,y)^{1/n}\}.$$ It's a closed subset of $X$. We assume that $d(x,y)\leqslant 1$ for all $x,y$, WLOG. Let $f\in V$, and $C,\alpha$ associated to this $f$. Take $n$ such that $n\geqslant C$ and $\frac 1n<\alpha$ to get that $f\in F_n$. Indeed, we have $$|f(x)-f(y)|\leqslant C\cdot d(x,y)^\alpha\leqslant n\cdot d(x,y)^\alpha= n\exp\left(\alpha\log\left(d(x,y)\right)\right)\leqslant n\cdot d(x,y)^{1/n}.$$

By Baire's theorem, we have that $F_{n_0}$ has a non-empty interior for some $n_0$, that is, exist, $f_0\in F_{n_0}$ and $r>0$ such that if $\lVert f-f_0\rVert_{\infty}\leq r$ then $f\in F_n$. For $f\in V$, we have $f_0+\frac r{2(1+\lVert f\rVert_{\infty})}f\in F_{n_0}$, hence $$|f(x)-f(y)|\leqslant |f_0(x)-f_0(y)|+\frac{2(1+\lVert f\rVert_{\infty})}r\cdot n_0d(x,y)^{1/n_0}\\ \leqslant n_0\left(1++\frac{2(1+\lVert f\rVert_{\infty})}r\right)d(x,y)^{1/n_0}.$$ Now, we can see that the unit ball of $V$ has a compact closure using Arzelà-Ascoli's theorem.

An other application can be found here.

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There is a partial differential equation with no solutions. Specifically, a first-order PDE on $\mathbb{R} \times \mathbb{C}$ with smooth coefficients, of the form $$\frac{\partial u }{\partial \bar{z}} - i z \frac{\partial u}{\partial t} = F(t,z).$$

See Lewy's example. I don't have the proof in front of me, but as I recall it goes by showing that the collection of $F$ for which a solution exists is meager in some appropriate space.

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I don't know if it is my favourite, but it is one of the few I know.

THM Let $(M,d)$ be a complete metric space with no isolated points. Then $(M,d)$ is uncountable.

PROOF Assume $M$ is countable, and let $\{x_1,x_2,x_3,\dots\}$ be an enumeration of $M$. Since each singleton is closed, each $X_i=X\smallsetminus \{x_i\}$ is open for each $i$. Moreover, each of them is dense, since each point is an accumulation point of $X$. By Baire's Theorem, $\displaystyle\bigcap_{i\in\Bbb N} X_i$ must be dense, hence nonempty, but it is readily seen it is empty, which is absurd. $\blacktriangle$.

COROLLARY Let $(M,d)$ be complete, $P$ a perfect subset of $M$. Then $P$ is uncountable.

PROOF $(P,d\mid_P)$ is a complete metric space with no isolated points.

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The Casorati–Weierstrass theorem says that if $G$ is an open subset of $\mathbb C$, $a$ is in $G$, and $f:G\setminus\{a\}\to\mathbb C$ is a holomorphic function such that $\lim\limits_{z\to a}f(z)$ does not exist in $\mathbb C\cup\{\infty\}$, then for each disk $D$ centered at $a$ (and contained in $G$), $f(D\setminus\{a\})$ is a dense subset of $\mathbb C$.

Baire's theorem can be used to give a strengthening of this result with the same hypotheses (still vastly weaker than Picard, but easier to prove). There exists a dense subset $X$ of $\mathbb C$ such that for each disk $D$ centered at $a$ (and contained in $G$), $f(D\setminus\{a\})$ contains $X$. In particular, this implies that for each $x\in X$, there is a sequence $(z_n)$ in $G\setminus\{a\}$ converging to $a$ such that $f(z_n)=x$ for all $n$.

To prove this, let $X=\bigcap\limits_{n=1}^\infty f\{z\in G:0<|z-a|<\frac{1}{n}\}$, note that each set in the intersection is open and dense by the open mapping theorem and the Casorati–Weierstrass theorem, apply Baire's theorem to see that $X$ is dense, and check that $X$ has the desired property.

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For any $1<p<\infty$ $$\bigcup_{q<p}\ell^q \not=\ell^p.$$

To see this note that $\ell^q$ is meagre in $\ell^p$ with respect to $\lVert\cdot\lVert_p$ since $\ell^q=\bigcup_{n\in\mathbb{N}}\{x\in\ell^q~|~\lVert x\lVert^q_q\leq n\}$.

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Here is a nice example I recently discussed in lecture.

Recall that a function $ f:\mathbb R\to \mathbb R $ is Baire one iff it is the pointwise limit of a sequence of continuous functions. These functions do not need to be continuous, but Baire proved that they always have uncountably many points of continuity; in fact, the set of points of continuity of $ f $ is a comeager $ G_\delta $ set.

To prove this, one first argues that all preimages $f^{-1}(\mathcal O) $ of open sets are $ F_\sigma $ sets (countable unions of closed sets). Now, $ f $ is not continuous at a point $ x $ iff there is an open interval $ I $ with rational endpoints and such that $ f (x)\in I $ but there is a sequence of points $ y $ converging to $ x $ with $ f (y)\notin I $, that is, $$ x\in f^{-1}(I)\cap \overline {\mathbb R\setminus f^{-1}(I)}. $$ Note that this is a meager $ F_\sigma $ set, and that therefore the set of points of discontinuity of $ f $ is a countable union of such sets, thus also a meager $ F_\sigma $ set.

It follows from this and the Baire category theorem that, in fact, the restriction of $ f $ to any nonempty closed set has a continuity point.

(Baire went on to show that the last statement is actually equivalent to $ f $ being Baire one.)

Since one can readily check that derivatives are Baire one, it follows that derivatives have many points of continuity.

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Another application of the Baire category theorem is to proof the Niemytzki Plane is not normal. It is a classical method, which called category method. see here: Application of Baire category theorem in Moore plane

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For a proof see page 10, example 2.22, here. – Rudy the Reindeer Jul 3 '12 at 6:43
Link (taken from comments to question): – sdcvvc Jul 4 '12 at 12:03

One of my favorite (albeit elementary) applications is showing that $\mathbb{Q}$ is not a $G_{\delta}$ set.

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My favorite application of Baire is due to Daniel Fischer - thanks a lot to Daniel!!!
(See his Answer to Uncountable Lebesgue Null Set.)

There exist uncountable Lebesgue null sets over the real line.
(These are precisely the ones giving rise to those obscure singular continuous measures.)

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I found one beautiful application of Baire Category Theorem which is the following:

Let $\mathcal H$ be a separable Hilbert Space with countable orthonormal basis $\{u_{k}\}_{k=1}^{\infty}$. Fix $n\in \mathbb N$ consider $\mathrm{Span}\{u_{1},u_{2},...,u_{n}\}$ then the following sets are dense in $\mathcal H$.

$A_{i,j}:=\{u\in \mathcal H: (u,u_{i})\neq (u,u_{j})\}$ where $1\leq i,j \leq n$ and $i\neq j$.

Proof Hints:(a) Any proper closed vector subspace of an Hilbert Space is nowhere dense. (b)A closed set is nowhere dense $\Leftrightarrow$ its complement is everywhere dense.

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You mean to say that any closed proper subspace is nowhere dense. – Asaf Karagila Jul 4 '12 at 7:51
@ Asaf Karagila: Yes I edited – users31526 Jul 4 '12 at 7:56
I don't understand. What you denote by $A_{i,j}$ is a finite intersection of open and dense sets: $$A_{i,j} = \bigcap_{1 \leq i \lt j \leq n} \mathcal{H} \smallsetminus (u_i - u_j)^\perp$$ which is obviously open and dense in $\mathcal{H}$ (no need for Baire here). And: what is the Span doing here? Are you intending to take a further countable intersection over $n$? Could you please clarify? – t.b. Jul 4 '12 at 11:24

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