Your favourite application of the Baire Category Theorem

Question

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:

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(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. $\Box$

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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.

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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."

Answer 1

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 $G_\delta$.

Answer 2

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 discussion in Math Overflow.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

$\overline{\mathbb Q_p}$ 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.

Answer 8

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

Answer 9

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

Answer 10

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.

Proof. By BCT, $C[0,1]$ under the supremum norm is a Baire space. For $p<q$ rational, the set $U_{p,q}$ of elements of $C[0,1]$ that are not monotone on $[p,q]$ is open and dense (if you are close to a wiggle, you must also wiggle, and you can always create a tiny wiggle if there is not already one). The intersection $U$ over all rationals $p<q$ is still dense by BCT, and therefore it is not empty. But an element of $U$ is not monotone on any interval of positive length since any such interval has a nontrivial subinterval with rational endpoints.

Answer 11

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.

Answer 12

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

Answer 13

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.

Answer 14

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 $$F_n:=\bigcap_{x,y\in [0,1]}\left\{f\in V,|f(x)-f(y)|\leqslant n\cdot d(x,y)^{1/n}\right\}.$$ Then $F_n$ is a closed subset of $V$. Indeed, it suffices to notice that for any fixed $x,y\in[0,1]$, the set $\left\{f\in V,|f(x)-f(y)|\leqslant n\cdot d(x,y)^{1/n}\right\}$ is a closed subset of $V$. This can be done in the following way: let $\left(f_l\right)_{l\geqslant 1}$ be a sequence of elements of $\left\{h\in V,|h(x)-h(y)|\leqslant n\cdot d(x,y)^{1/n}\right\}$ which converges uniformly on $X$ to $f$. Then for each $l$, $\left|f_l(x)-f_l(y)\right|\leqslant n d(x,y)^{1/n}$ and taking the limit $l\to +\infty$, we derive that $\left|f (x)-f (y)\right|\leqslant n d(x,y)^{1/n}$.

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.

Answer 15

Here's a geometric group theory application.

Let $G$ be a finitely generated group. Let $T$ be a real tree on which $G$ acts by isometries, in such a way that there does not exist a $G$-invariant proper subtree. Suppose that branch points are dense in $T$; equivalently, there exists no open subset of $T$ that is an embedded open arc, i.e. is homeomorphic to $(0,1)$. Let $G$ act on $T$ by isometries so that the action is minimal, meaning there exists no proper $G$-invariant subtree of $T$. Then $T$ is not complete.

The proof is that by using finite generation of $G$ and minimality of the action, one can find an embedded closed arc $\alpha \subset T$, meaning $\alpha$ is homemorphic to $[0,1]$, such that its set of translates $\{g \cdot [0,1] \mid g \in G\}$ covers $T$. But then, since branch points are dense, the arc $\alpha$, and each of its translates, is nowhere dense. Thus $T$ is covered by countably many nowhere dense sets.

This failure of completeness effects several arguments which use group actions on real trees, for example the proof by Levitt and Lustig that an irreducible outer automorphisms of the rank $n$ free group $F_n$ acts with north--south dynamics on the compactified outer space of $F_n$.

Answer 16

There are a few more techniques than just BCT going on here, but BCT finishes things off. Suppose $B(\mathbb D)$ is the ring of bounded holomorphic functions on the unit disc $\mathbb D = \{ z \in \mathbb C : |z| < 1 \}$.

Theorem. There is no Polish topology on $B(\mathbb D)$ that makes $B(\mathbb D)$ into a Polish ring (Polish topology so that all of the ring operations are continuous).

Assume $B(\mathbb D)$ is a Polish ring, let $H^\infty$ be the set of bounded holomorphic functions on the disk with the uniform convergence topology (or the sup-norm topology), and let $\varphi : B(\mathbb D) \to H^\infty$ be the identity. Effectively, one uses an algebraic characterization of the constant functions due to H.L. Royden and the ideal structure to show eventually that $f \mapsto f(\alpha)$, $B(\mathbb D) \to \varphi^{-1}[\mathbb C]$, is continuous. Then it follows that $F_n := \{ f \in B(\mathbb D) : \|f\|_\infty \leq n\}$ are all closed subsets of $B(\mathbb D)$ so, since we assumed it was Polish, you can apply BCT to find one of those with non-empty interior which then further implies that $\varphi$ is continuous. But the continuous image of a separable space is separable and $H^\infty$ is not separable, so we have a contradiction.

Answer 17

Another interesting application : Dimension of a Banach space can't be countably infinite i.e $\dim(X) \neq \aleph_{o}$.

Proof: $(X, \|•\|) $ be a Banach space.

Suppose $\dim(X) =\aleph_o$ And $\mathcal{B}=\{e_1,e_2,\ldots\}$ be a Hamel basis.

Let,$$E_1=span\{e_1\}$$ $$E_2=span\{e_1,e_2\}$$

$$\dots$$

$$E_n=span\{e_1,e_2,\ldots,e_n\}$$

Then $X=\bigcup_{n\in\Bbb{N}}E_n$

Where each $E_n$ is proper closed subspace , hence n.w.dense .

Hence, $X$ is of first category. It's a contradiction as any Banach space is of second category (Baire's theorem)

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Any finite dimensional subspace of a normed space is closed

Any proper subspace of a normed space must have empty interior i.e contains no ball.