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

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 2

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 3

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 4

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 5

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

Answer 6

$\overline{\mathbb Q_p}$ 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 7

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 8

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.

Answer 9

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 10

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

Answer 11

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.

Answer 12

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 13

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

Answer 14

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.

Answer 15

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 16

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

Answer 17

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.

Answer 18

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 19

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

Answer 20

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.

Answer 21

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

Answer 22

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

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. $\blacksquare$

Answer 23

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.

Answer 24

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.)

Answer 25

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).

Answer 26

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 27

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 28

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 $a<b$, 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 $$ \liminf_{x\to\infty}f(x)\ge (j-1)\varepsilon \quad\text{while}\quad \limsup_{x\to\infty}f(x)\le (j+1)\varepsilon. $$

Answer 29

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.

Answer 30

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.