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For integers $m,n\ge 1$, let $E_{m,n}$ contain those $f\in C([0,1])$ for which there is some $x\in [0,1]$ such thatwhenever $|y-x|<1/n$. Compactness of $[0,1]$ shows that $E_{m,n}$ is closed in the uniform norm: for a uniformly convergent sequence, choose convergent subsequences of the corresponding witness points and pass to the limit.

Each $E_{m,n}$ has empty interior. Given $f$ and a uniform-error tolerance, first approximate $f$ by a polygonal function, then add a sufficiently small-amplitude, sufficiently high-frequency sawtooth whose slopes dominate both $m$ and every slope of the polygonal approximation. At every point, one can move a distance below $1/n$ along one adjacent linear segment and obtain a difference quotient of magnitude greater than $m$. The perturbed function lies in the prescribed ball but outside $E_{m,n}$.

If $f$ is differentiable at some point, its nearby difference quotients are bounded, so $f\in E_{m,n}$ for some $m,n$. The functions differentiable somewhere therefore form a meagre set. By the Baire category theorem, their complement is nonempty; indeed it is dense. Thus

Write $C(K)={\bigcup }_{n\ge 1}{\mathcal{S}}_n$ with every ${\mathcal{S}}_n$ equicontinuous. Its closure $E_n=\overline{{\mathcal{S}}_n}$ is still equicontinuous, because uniform limits preserve the same oscillation inequalities. The Baire category theorem applied to the Banach space $C(K)$ shows that some $E_n$ has nonempty interior. Hence there are $f_0\in C(K)$ and $r>0$ such that the open ball $B(f_0,r)$ lies in $E_n$.

Fix $x\in K$. Equicontinuity of $E_n$ gives a neighborhood $U$ of $x$ on which every member of $E_n$ changes by less than $r/4$. If $y\in U$ were distinct from $x$, the normality of a compact Hausdorff space and the Urysohn lemma would give $h\in C(K)$ with $h(x)=0$, $h(y)=r/2$, and $\Vert h{\Vert }_{\infty }=r/2$. Both $f_0$ and $f_0+h$ belong to $E_n$, yeta contradiction. Thus $U=\{x\}$, so every point is isolated and $K$ is a discrete space. A compact discrete space is finite, since its cover by singleton open sets has a finite subcover.

A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

If a subspace $W\subset V$ contains a ball about some $w\in W$, translation gives a ball about zero in $W$, and scaling then gives every vector of $V$. Thus every proper subspace has empty interior.

If a complete infinite-dimensional $V$ had a countable Hamel basis $(e_n)$, thenEach finite-dimensional subspace is closed and, being proper, nowhere dense. This contradicts the Baire category theorem. The polynomials have the countable Hamel basis $1,x,x^2,\dots$ and are infinite-dimensional, so no norm can make that whole vector space complete.

False. Fix $y_0\in \mathbb{R}$ and takeThis closed set has empty interior in ${\mathbb{R}}^2$, so it is nowhere dense and hence meagre. Its section at $y_0$ is $C_{y_0}=\mathbb{R}$, which is not meagre in itself by the Baire category theorem.

A set is of first Baire category, or meagre, when it is a countable union of nowhere dense sets. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant Baire category theorem says that a nonempty complete metric space is not meagre in itself.

False. The coordinate axesare both closed nowhere dense subsets of ${\mathbb{R}}^2$, but every $(x,y)$ is $(x,0)+(0,y)$, sowhich is not meagre by the Baire category theorem.

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

For rational $q\in (0,1)$ and positive integer $M$, defineEvaluation of the $m$th derivative at $q$ is continuous in the smooth-function metric, so $U_{q,M}$ is open.

It is also dense. Given $f$ and a metric tolerance $\epsilon >0$, choose $R\ge 1$ so that the contribution of all derivatives of order at least $R$ is below $\epsilon /2$, and choose $m\ge \max \{M,R\}$. For large $K$, perturb $f$ byFor every $r<R$, $\Vert h_K^{(r)}{\Vert }_{\infty }\le \epsilon /(4R)$, so $d(f+h_K,f)<\epsilon$. On the other hand,For sufficiently large $K$, $|f^{(m)}(q)+h_K^{(m)}(q)|>m!m^m$, proving density.

The space is complete by part (b). The Baire category theorem therefore makesdense. Put $E=C^{\infty }([0,1])\setminus G$. Each complement $U_{q,M}^c$ is closed and nowhere dense, so $E$ is a meagre set, or a set of first category. Every $f\notin E$ has the required derivative growth. This is the generic superfactorial derivative growth at rational points.

The Baire category theorem states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\dots$ be open dense sets and start with any nonempty open set $V$. Inductively choose closed ballswith the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense, which is the equivalent form of the theorem.

Now each $nS$ is closed and ${\bigcup }_{n\ge 1}nS=X$. Baire givesfor some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\Vert z\Vert <r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hencewhich proves the closed convex absorbing set has an origin neighbourhood result.

Convexity cannot be dropped. In $X=\mathbb{R}$, setThis is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne 0$, choose $k$ so that the intervalhas length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so ${\bigcup }_{n}nS=\mathbb{R}$, while $S$ is not a neighbourhood of zero. This is a closed symmetric absorbing set without an origin neighbourhood.

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere-dense closed sets. Equivalently, every countable intersection of open dense sets is dense.

A point $x$ of a metric space is an isolated point when some $\epsilon >0$ satisfies $B(x,\epsilon )=\{x\}$.

True. If a nonempty countable complete metric space had no isolated point, every singleton would be closed and nowhere dense. Their countable union would be the whole space, contradicting the Baire category theorem. This is the basic isolated points in a countable complete metric space argument.

Part i makes $K_{\epsilon }^{\prime }$ a closed subset of the compact metrizable space $K$, so it is compact metrizable and therefore separable. Choose the stated dense sequence $(f_m)$ and, using part iii, the sequences $(g_{m,n})$ and $(y_{m,n})$. In constructing the universal weakly null sequence $(z_n)$ from part b, retain a tail of each $y_m$ and write $z_{j(m,n)}=y_{m,n}$ along the retained subsequence, where $j(m,n)\to \infty$.

Eachis weak-star closed. Since $z_n\rightharpoonup 0$, every fixed $f\in K$ belongs to some $A_N$, so $K={\bigcup }_N{A}_N$. The compact Hausdorff space $K$ is a Baire space, and the Baire category theorem implies that some $A_N$ has nonempty relative weak-star interior.

Suppose $K_{\epsilon }^{\prime }=K$. Choose a nonempty relatively open $O\subseteq A_N$ and then $f_m\in O$ by density. Since $g_{m,n}\stackrel{w^{\ast }}{\to }{f}_m$, eventually $g_{m,n}\in O$. For a sufficiently late retained index, also $j(m,n)\ge N$, and hencecontradicting $|g_{m,n}(y_{m,n})|>\epsilon /16$. Therefore the One-step Szlenk derivation for a separable dual gives $K_{\epsilon }^{\prime }⊊K$ whenever $K$ is nonempty.

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.

Let $C=C([0,1])$ with the uniform norm. For rational numbers $0\le a<b\le 1$, let $M_{a,b}^{+}$ and $M_{a,b}^{-}$ be the sets of functions that are respectively nondecreasing and nonincreasing on $[a,b]$.

Both sets are closed. For example, if $f_j\in M_{a,b}^{+}$ and $f_j\to f$ uniformly, then for $x<y$ in $[a,b]$,They also have empty interior. Given $f\in M_{a,b}^{+}$ and $\epsilon >0$, continuity lets us choose $a<x<y<b$ sufficiently close thatAdd a continuous triangular bump $h$ supported near $x$, with $h(x)=3\epsilon /4$, $h(y)=0$, and $\Vert h{\Vert }_{\infty }<\epsilon$. Then $(f+h)(x)>(f+h)(y)$, so the $\epsilon$-ball about $f$ is not contained in $M_{a,b}^{+}$. The analogous upward bump at $y$ deals with $M_{a,b}^{-}$. Thus both sets are nowhere dense.

There are only countably many rational pairs $(a,b)$, sois meagre. Completeness of $C$ and the Baire category theorem show that its complement is nonempty. Choose $f$ in that complement. If $f$ were monotone on an interval of positive length, that interval would contain a closed interval with rational endpoints, putting $f$ in one of the displayed sets. Hence $f$ is monotone on no interval of positive length, as described by the generic nowhere-monotone continuous function result.

A subset $A$ of a metric space $X$ is a nowhere dense set whenEquivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

The Baire category theorem says that if $X$ is a complete metric space and $G_1,G_2,\dots$ are open dense subsets of $X$, then ${\bigcap }_{n\ge 1}{G}_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.

To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ballwith $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to chooseThe balls are nested, and for $m>n$ their centres satisfyThus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. HenceAs $V$ was arbitrary, the intersection is dense.

A subset is dense when every nonempty open set meets it, equivalently its closure is the whole space. Baire category theorem says a complete metric space has dense intersection for every countable family of open dense sets. Hereso Baire makes $S$ dense. It need not be open: enumerate the rationals $q_j$ and take $A_j=[0,1]\setminus \{q_1,\dots ,q_j\}$, giving the irrationals. Nor need ${\cap }_j{A}_j$ be nonempty: take $A_1=\varnothing$ and $A_j=[0,1]$ for $j\ge 2$.

Suppose that for every $z\in \mathbb{C}$, at least one derivative $f^{(n)}(z)$ of an entire function vanishes. The closed sets $E_n=\{z:f^{(n)}(z)=0\}$ cover $\mathbb{C}$. By the Baire category theorem, some $E_N$ has nonempty interior. The identity theorem gives $f^{(N)}=0$ everywhere, so $f$ is a polynomial of degree below $N$.

On a compact smooth manifold without boundary of dimension at least two, a residual set of smooth Riemannian metrics consists of nowhere locally homogeneous metrics. Thus such metrics are dense in the smooth topology by the Baire category theorem. In dimension one, arclength coordinates always provide nontrivial local isometries.