Dtyjlui

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.

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.

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.

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

$overline{mathbb Q_p}$ 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.

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.

The rationals are not completely metrizable.

Proof: Since the rationals have no isolated points, $mathbb Qsetminus{q}$ is dense and open for every $q$, but $bigcap_{qinmathbb Q}mathbb Qsetminus{q}$ 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 $G_delta$.

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.

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.

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

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 $X_i=Xsmallsetminus {x_i}$ is open for each $i$. Moreover, each of them is dense, since each point is an accumulation point of $X$. By Baire's Theorem, $displaystylebigcap_{iinBbb N} X_i$ 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 $(P,dmid_P)$ is a complete metric space with no isolated points.

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

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 $fin V$, we can find $C>0$ and $0<alpha< 1$ such that
$$forall x,yin X,quad |f(x)-f(y)|leqslant Ccdot d(x,y)^{alpha}.$$
Then $V$ is finite dimensional.

Define
$$F_n:=bigcap_{x,yin [0,1]}left{fin V,|f(x)-f(y)|leqslant ncdot d(x,y)^{1/n}right}.$$
Then $F_n$ is a closed subset of $V$. Indeed, it suffices to notice that for any fixed $x,yin[0,1]$, the set $left{fin V,|f(x)-f(y)|leqslant ncdot d(x,y)^{1/n}right}$ is a closed subset of $V$. This can be done in the following way: let $left(f_lright)_{lgeqslant 1}$ be a sequence of elements of $left{hin V,|h(x)-h(y)|leqslant ncdot d(x,y)^{1/n}right}$ 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 $lto +infty$, 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 $fin V$, and $C,alpha$ associated to this $f$. Take $n$ such that $ngeqslant C$ and $frac 1n<alpha$ to get that $fin F_n$. Indeed, we have
$$|f(x)-f(y)|leqslant Ccdot d(x,y)^alphaleqslant ncdot d(x,y)^alpha=
nexpleft(alphalogleft(d(x,y)right)right)leqslant ncdot d(x,y)^{1/n}.$$

By Baire's theorem, we have that $F_{n_0}$ has a non-empty interior for some $n_0$, that is, exist, $f_0in F_{n_0}$ and $r>0$ such that if $lVert f-f_0rVert_{infty}leq r$ then $fin F_n$. For $fin V$, we have
$f_0+frac r{2(1+lVert frVert_{infty})}fin F_{n_0}$, hence
$$|f(x)-f(y)|leqslant |f_0(x)-f_0(y)|+frac{2(1+lVert frVert_{infty})}rcdot n_0d(x,y)^{1/n_0}\
leqslant n_0left(1++frac{2(1+lVert frVert_{infty})}rright)d(x,y)^{1/n_0}.$$
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.

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 $lVertcdotlVert_p$ since $ell^q=bigcup_{ninmathbb{N}}{xinell^q~|~lVert xlVert^q_qleq n}$.

Here is a nice example I recently discussed in lecture.

Recall that a function $ f:mathbb Rto 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, $$ xin f^{-1}(I)cap overline {mathbb Rsetminus f^{-1}(I)}. $$
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.

The Casorati–Weierstrass theorem says that if $G$ is an open subset of $mathbb C$, $a$ is in $G$, and $f:Gsetminus{a}tomathbb C$ is a holomorphic function such that $limlimits_{zto a}f(z)$ does not exist in $mathbb Ccup{infty}$, then for each disk $D$ centered at $a$ (and contained in $G$), $f(Dsetminus{a})$ 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 $G$), $f(Dsetminus{a})$ contains $X$. In particular, this implies that for each $xin X$, there is a sequence $(z_n)$ in $Gsetminus{a}$ converging to $a$ such that $f(z_n)=x$ for all $n$.

To prove this, let $X=bigcaplimits_{n=1}^infty f{zin G:0<|z-a|<frac{1}{n}}$, 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.

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

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

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.)

Baires category theorem can be very useful if you consider CW-complexes, especially if you want to prove that a space does not admit a CW-structure (in connection with considering Baire spaces).

For example an infinite-dimensional Hilbert space is not a CW-complex since it is a Baire space (see i.e. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable. ). Another example with a similar but extended argument can be found here https://mathoverflow.net/questions/152802/all-mapping-space-between-cw-complexes-is-a-cw-complex, it is about the cellular mapping space $Map(X,Y)$ for $X$ and $Y$ finite CW-complexes.
And there are of course many other of such examples.

Another less known but important application of the Baire category theorem is in the proof of the Vitali-Hahn-Saks theorem. The proof can be found in these posts
finite measure case and infinite measure case

Let $f:mathbb R^+to mathbb R$ be continuous. Suppose that for all $x>0,$ $lim_{ntoinfty} f(nx)=0$. Then $lim_{xto infty} f(x)=0$.

Proof: Fixing $epsilon>0$, let $K_n={x:|f(mx)|leepsilontext{ for all }mge n$}. Then the $K_n$ are closed and their union is $mathbb R^+$, so BCT implies that some $K_n$ contains an interval, $[a,b]$. This implies that $|x|<epsilon$ for all $f(x)$ in the set $$[na,nb]cup [(n+1)a,(n+1)n]cup [(n+2)a,(n+2)b]cupdots.$$ You can show this union of intervals contains an interval of the form $[M,infty)$ for some $M$, so $|f(x)|<epsilon$ for large enough $x$, completing the proof.

$newcommand{wh}{widehat}$
$newcommand{set}[1]{{#1}}$
$newcommand{vp}{varphi}$

Theorem 1.
Let $M$ and $N$ be smooth manifolds and $F:Mto N$ be a surjective smooth map of constant rank.
Then $F$ is a smooth submersion.

Proof.
For each point $pin M$, we can find charts $(U_p,vp_p)$ and $(V_p,psi_p)$ containing $p$ and $f(p)$ respectively such that $overline{f(U_p)}subseteq V_p$ (here, just to be clear, the closure is taken in $N$).
In the light of the rank theorem, we can further assume without loss of generality that the following holds
$$
psi_pcirc fcirc vp_p^{-1}(x_1,ldots,x_k,x_{k+1},ldots,x_m)=(x_1,ldots,x_k,0,ldots,0),quad forall (x_1,ldots,x_m)in wh U_p
$$
We will denote $psi_pcirc fcirc vp_p^{-1}$ as $hat f_p$.
Note that $hat f_p(wh U_p)$ is nowhere dense in $wh V_p$.
Since $psi_p^{-1}:wh V_pto V_p$ is a homeomorphism, we conclude that $psi_p^{-1}(hat f_p(wh U_p))= f_p(U_p)$ is nowhere dense in $V_p$.
Since $overline{f_p(U_p)}subseteq V_p$, we infer that $f_p(U_p)$ is in fact nowhere dense in $N$.
Now since $M$ is second countable, we can find a countable subset $C$ of $M$ such that $set{U_p}_{pin C}$ covers $M$.
By surjectivity of $f$, we infer that $set{f_p(U_p)}_{pin C}$ covers $N$.
But this means that $N$ is a countable union of nowhere dense subsets.
Since $N$ is locally compact Hausdorff, this contradicts the fact that $N$ is a Baire space and we are done.
$blacksquare$