# Your favourite application of the Baire Category Theorem

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."

• mathoverflow.net/questions/34059/…
– user38268
Commented Jul 2, 2012 at 13:34
• Commented Jul 2, 2012 at 13:39
• 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. Commented Jul 2, 2012 at 22:27
• A nonempty complete metric space $X$ is not a countable union of nowhere dense sets. Commented Jul 3, 2012 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. Commented Apr 12, 2013 at 19:18

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.

• The other one of the big theorems from beginning functional analysis that uses the Baire Category Theorem is the open mapping theorem Commented Jul 2, 2012 at 14:05
• And its friend the closed graph theoerem. Commented Jul 2, 2012 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? Commented Jul 2, 2012 at 14:08
• @ MattN It looks like someone already did. Commented Jul 2, 2012 at 14:23
• @FrancisAdams Yes I saw. Btw, if you put a space in between "@" and my name I won't get pinged. Commented Jul 2, 2012 at 19:14

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.

• 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 $\mathcal C(I)$ Commented Aug 9, 2012 at 12:10
• math.stackexchange.com/questions/2993112/… Commented Feb 15, 2022 at 13:17

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.

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

• I'm surprised that you have only one favourite! :-) Commented Jul 3, 2012 at 9:24

$\overline{\mathbb Q_p}$ 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.

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 $G_\delta$.

• This one? Commented Jul 2, 2012 at 13:30
• What's long about "since it's the union of its countably many points and no point is open"? :)
– t.b.
Commented Jul 2, 2012 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! Commented Jul 2, 2012 at 14:12
• @AsafKaragila You could still add the proof... Commented Jul 2, 2012 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. Commented Jul 3, 2012 at 1:05

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.

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 $$p rational, the set $$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 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.

• Could you also provide the link with the proof of this very nice result?
– RFZ
Commented Nov 6, 2019 at 20:06
• Does this really require BCT? Any highly oscillatory function (such as Weierstrass's function) would do, no? Commented Aug 17, 2022 at 19:58
• @ZFR, I apologize I forget where this result is from but I've added a proof sketch. And Willie of course this result does not require BCT, but the BCT proof is quite simple and doesn't require any calculations, whereas constructing and analyzing a highly oscillatory function usually takes some careful thought. Of course, it is always nice to have highly oscillatory functions as concrete examples. Commented Nov 1, 2022 at 0:08

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

• Drats. I was gonna post that one when I got home, but I see I got here seven minutes too late! Damn buses! :-) Commented Jul 2, 2012 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. Commented Jul 3, 2012 at 23:44
• @Fëanor: I haven't read it. What is the title and where can I find it? Commented Jul 4, 2012 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 math.stackexchange.com/questions/146910/… Commented Jul 4, 2012 at 17:12

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.

• (1).Nitpicking: Assuming M is not the empty space. (2). Without Baire we can show by elementary means that a complete metric space $M$ has a subspace homeomorphic to the Cantor set, so, regardless of the Continuum Hypothesis, the cardinal of $M$ is at least $2^{\aleph_0}.$ Commented Dec 16, 2017 at 10:45

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.

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

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.

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.

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 $\ell^q=\bigcup_{n\in\mathbb{N}}\{x\in\ell^q~|~\lVert x\lVert^q_q\leq n\}$.

Let $$f:\mathbb R^+\to \mathbb R$$ be continuous. Suppose that for all $$x>0,$$ the discrete sequence $$f(x),f(2x),f(3x),\dots,f(nx),\dots$$converges to zero. Then $$\lim_{x\to \infty} f(x)=0$$.

Proof: Fixing $$\epsilon>0$$, let $$K_n=\{x:|f(mx)|\le\epsilon\text{ for all }m\ge n$$}. 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 $$|f(x)|<\epsilon$$ for all $$f(x)$$ in the set $$[na,nb]\cup [(n+1)a,(n+1)b]\cup [(n+2)a,(n+2)b]\cup\dots.$$ 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.

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

• For a proof see page 10, example 2.22, here. Commented Jul 3, 2012 at 6:43
• Link (taken from comments to question): math.stackexchange.com/questions/135947 Commented Jul 4, 2012 at 12:03
• Interesting. This can also be answered by the "lowest-order" case of the Jones Lemma: A separable Tychonoff space with a closed discrete subspace of cardinal $2^{\aleph_0}$ is not normal. Commented Dec 16, 2017 at 10:53

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

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

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

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

Proof. For each point $$p\in M$$, 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 $$\psi_p\circ f\circ \vp_p^{-1}(x_1,\ldots,x_k,x_{k+1},\ldots,x_m)=(x_1,\ldots,x_k,0,\ldots,0),\quad \forall (x_1,\ldots,x_m)\in \wh U_p$$ We will denote $$\psi_p\circ f\circ \vp_p^{-1}$$ as $$\hat f_p$$. Note that $$\hat f_p(\wh U_p)$$ is nowhere dense in $$\wh V_p$$. Since $$\psi_p^{-1}:\wh V_p\to V_p$$ 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 $$\set{U_p}_{p\in C}$$ covers $$M$$. By surjectivity of $$f$$, we infer that $$\set{f_p(U_p)}_{p\in C}$$ 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. $$\blacksquare$$

Another application of Baire's Theorem:

If $$f:\mathbb R\setminus\mathbb Q\to\mathbb Q$$ is continuous, then there exists an open interval $$I=(a,b)$$, such that $$f$$ restricted in $$I\cap(\mathbb R\setminus\mathbb Q)$$ is constant. If fact, there exists a family of disjoint open intervals $$\{I_n\}$$, whose union is dense in $$\mathbb R$$, such that when $$f$$ is restricted in $$I_n\cap(\mathbb R\setminus\mathbb Q)$$ is constant, for every $$n$$.

Proof. Let $$\{q_n\}$$ be an enumeration of $$\mathbb Q$$ and $$A_n=f^{-1}[\{q_n\}]$$, $$n\in\mathbb N$$. Then the $$A_n$$'s are closed in $$\mathbb R\setminus\mathbb Q$$ and hence there exist $$F_n$$ closed in $$\mathbb R$$, such that $$A_n=F_n\cap(\mathbb R\setminus\mathbb Q)$$. Clearly, $$\mathbb R=\bigcup_{n\in\mathbb N}F_n\cup\mathbb Q= \bigcup_{n\in\mathbb N}F_n\cup\bigcup_{n\in\mathbb N}\{q_n\}.$$ Hence, there is an $$F_n$$ with $$\mathrm{int}(F_n)\ne\varnothing$$. In order to obtain the family of $$I_n$$'s, we repeat this argument replacing $$f$$ by its restriction in $$I\cap (\mathbb R\setminus\mathbb Q)$$, where $$I$$ is an arbitrary open interval.

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.)

• Isn't this much more easily proved by just considering the Cantor set? (I think you also want the set to be comeager, which is more interesting.) Commented Nov 1, 2022 at 1:42

If $$X$$ is a compact metric n-dimensional space, then the set of embeddings of X to $$\mathbb{R}^{2n+1}$$ is dense in $$C(X,\mathbb{R}^{2n+1})$$ (with respect to sup-norm).

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 $$F_n$$.

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.

Fact. Let $$f:(0,\infty)\to\mathbb{R}$$ be a continuous function, and suppose that $$\lim_{n\to\infty} f(nx)$$ exists in $$\mathbb R$$, for all $$x>0$$. Then $$\lim_{x\to\infty} f(x),$$ exists as well.

Proof. (Using Baire's Theorem.), Our target is to prove that the difference $$\limsup_{x\to\infty}f(x)-\liminf_{x\to\infty}f(x)$$ is arbitrarily small. Let $$\varepsilon>0$$. We define $$F_{n,j}=\big\{x\in (0,\infty) : |f(mx)-j\varepsilon|\le\varepsilon, \quad\text{for all m\ge n}\big\}, \quad n\in\mathbb N,\,\, j\in\mathbb Z.$$ Clearly, $$\bigcup_{j\in \mathbb Z}\bigcup_{n\in \mathbb N}F_{n,j}=(0,\infty).$$ Baire's Theorem provides that at least one the $$F_{n,j}$$'s has nonempty interior. In particular, there exist $$n\in\mathbb N$$, $$j\in\mathbb Z$$, and $$a,b>0$$, with $$a, such that $$[a,b]\subset F_{n,j}.$$ However, $$[a,b]\subset F_{n,j},$$ implies that $$[ka,kb]\subset F_{n,j},$$ for all $$k\in\mathbb N$$, i.e., $$[a,b]\cup[2a,2b]\cup\cdots\cup [ka,kb]\cup\cdots\subset F_{n,j}.$$ Next observe that if $$k$$ is sufficiently large so that $$kb>(k+1)a$$ which happens for $$k>a/(b-a)$$, then $$[ka,kb]\cup [(k+1)a,(k+1)b]\cup\cdots = [ka,\infty),$$ and thus $$[ka,\infty)\subset F_{n,j}=\big\{x\in (0,\infty) : |f(mx)-j\varepsilon|\le\varepsilon, \quad\text{for all m\ge n}\big\}.$$ This implies that $$[nka,\infty)\subset \big\{x\in (0,\infty) : |f(x)-j\varepsilon|\le\varepsilon\big\}$$, and thus $$x\ge nka \quad\Longrightarrow\quad (j-1)\varepsilon \le f(x)\le (j+1)\varepsilon.$$ Therefore $$\liminf_{x\to\infty}f(x)\ge (j-1)\varepsilon \quad\text{while}\quad \limsup_{x\to\infty}f(x)\le (j+1)\varepsilon.$$

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)

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.

• This one was already mentioned above Commented Nov 19, 2023 at 11:37

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.

• You mean to say that any closed proper subspace is nowhere dense. Commented Jul 4, 2012 at 7:51
• @ Asaf Karagila: Yes I edited Commented Jul 4, 2012 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.
Commented Jul 4, 2012 at 11:24