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Every proper affine hyperplane in ${\mathbb{R}}^n$ is closed and has empty interior: a small displacement along its nonzero normal vector leaves it. The Baire category theorem therefore makes the complement of a countable collection of such hyperplanes dense. In particular, avoiding the hyperplanes ${\sum }_{j=1}^n{k}_j{x}_j=k_{n+1}$ for nonzero integer tuples yields a vector for which $1,x_1,\dots ,x_n$ are linearly independent over $\mathbb{Q}$. Tuples with all the first $n$ entries zero define an empty forbidden set.

No infinite-dimensional Banach space has a countable Hamel basis. If $(e_n)$ were such a basis, the space would be the countable union of the finite-dimensional subspaces $\mathrm{span}(e_1,\dots ,e_n)$. Each is closed and has empty interior, contradicting the Baire category theorem.

Write the dense space $X$ as ${\bigcap }_n{G}_n$ in compact Hausdorff $K$. Every $G_n$ is dense and open. A dense open subset $U_n$ of $X$ has the form $X\cap O_n$ with $O_n$ dense and open in $K$. The compact Baire category theorem makes ${\bigcap }_n(G_n\cap O_n)={\bigcap }_n{U}_n$ dense in $K$, hence in $X$. Therefore every Čech-complete space is a Baire space. The rationals are not a Baire space, as deleting their points one at a time gives dense open sets with empty intersection.

The complement of a comeagre set is a meagre set, a countable union of nowhere dense sets. A dense G-delta set is comeagre. The Baire category theorem ensures that two comeagre subsets of a nonempty complete metric space cannot be disjoint.

Suppose a linear map $S:c_0\to Y$ has compact image of the closed unit ball. It is bounded, and each $g\in Y^{\ast }$ attains its maximum on this compact image. Thus $S^{\ast }g$ attains its norm, so norm attainment on the vanishing sequence space implies $S^{\ast }(Y^{\ast })\subseteq c_{00}$. The Baire category theorem applied to the inverse images of the first $n$ coordinate spans makes this adjoint finite rank: these are closed linear subspaces covering the complete continuous dual space $Y^{\ast }$, so one has interior and is all of $Y^{\ast }$. Finally the Hahn-Banach theorem identifies $\ker S$ as the common kernel of $S^{\ast }(Y^{\ast })$, proving finite rank of $S$. Ordinary compact operators require only relative compactness of the ball image, so they need not have finite rank.

Choose a compact $K\subset [0,1]\times \mathbb{R}$ with full horizontal projection and zero-length projections $y+tx$ for $t\in [0,1]$. Then the displayed set is compact, has zero area by the Fubini theorem, and contains a segment of every slope in $[0,1]$. Such parameter sets exist by the Baire category theorem in the Hausdorff metric: fine parallel segments approximate any parameter set while making a fixed band of projections uniformly small. Finite grids and a countable intersection make every required projection null. Four rotated copies supply every planar direction.

For countably many linearly independent vectors in an infinite-dimensional irreducible complex Banach module, there are elements $a_n$ satisfying the displayed annihilation conditions while the tail images $\{\pi (a_n\cdots a_1){\xi }_j:j\ge n\}$ remain linearly independent. Use the quotient norm on an irreducible Banach module to make each finite annihilator a closed Banach subspace. The Jacobson density theorem makes each condition of independence of finitely many tail images open and dense there; the Baire category theorem intersects these countably many conditions. This is stronger than prescribing only finitely many images once.

In a finite hyperbolic covering, primitive lifts of a primitive closed geodesic correspond to cycles of its monodromy permutation; an $r$-cycle gives length $r\ell (\gamma )$. In a real-analytic family of compact hyperbolic surfaces or hyperbolic cone orbifolds, a prescribed multiple of a simple curve's length can be isolated generically from unrelated lifted lengths. To exclude an identity of length functions, pinch the simple curve: only its peripheral powers can have length tending to zero, while intersecting curves are long by the collar lemma. The countably many remaining length equalities have empty interior, so the Baire category theorem supplies a parameter avoiding them. Only finitely many closed geodesics lie below a fixed local length bound; thus the isolation persists in an open parameter neighbourhood. This makes cycle-intersection information an intrinsic obstruction to an isometry, rather than an invariant of a chosen covering map alone.

In $C([0,1])$ with the uniform norm, the continuous functions that are monotone on some interval of positive length form a meagre set. Indeed, it is enough to use intervals with rational endpoints. For each such interval, the nondecreasing and nonincreasing functions form closed sets with empty interior: a sufficiently small local triangular perturbation breaks the relevant inequality. The Baire category theorem therefore shows that a dense set of continuous functions is monotone on no nontrivial interval.

Outside a meagre set in $C^{\infty }([0,1])$, every rational $q\in (0,1)$ and every positive integer $M$ admit $m\ge M$ such thatFor fixed $q,M$, the union over $m\ge M$ of these strict-inequality sets is open. It is dense because a perturbation $\delta \cos (K(x-q)-m\pi /2)$ can make the $m$th derivative arbitrarily large while keeping any prescribed finite collection of lower derivatives arbitrarily small. The Baire category theorem applied over the countable pairs $(q,M)$ gives the claim.

For each integer $m\ge 1$, the set of continuous functions satisfying $|S_N(f,0)|\le m$ for every $N$ is closed in the uniform norm. If it contained a ball of radius $r$, subtracting two members of that ball would bound every partial-sum functional norm by $4m/r$. The Dirichlet kernel harmonic lower bound excludes this, so these sets are nowhere dense. By the Baire category theorem, their complement is a dense $G_{\delta }$ set. Every function in that complement has unbounded Fourier partial sums at zero. Translation gives the same conclusion at any prescribed point. This is a direct Uniform boundedness principle argument on the real Banach space of continuous functions on the circle.

For five pairs of continuous functions on $[0,1]$, one can arrange that at every arbitrarily fine scale there are closed interval families whose product rectangles cover each point at least three times, and within each family the images under $a_q(x)+b_q(y)$ are mutually disjoint. To see this, put gaps near five shifted fine grids; gap sets for different shifts are disjoint. Each coordinate misses at most one family, so a pair misses at most two. Approximate any prescribed pair of functions by constants on the complementary intervals, interpolating across gaps. Perturb the finitely many constants to avoid all equalities between distinct pair sums. The rectangle images are then separated by a positive distance, a property preserved under sufficiently small uniform perturbations. For each required mesh size this property defines a dense open subset of $C([0,1]{)}^{10}$. The Baire category theorem gives a tuple satisfying every scale simultaneously.

Every nonempty countable complete metric space has an isolated point. Otherwise it is the countable union of its singleton subsets, each of which is nowhere dense, contradicting the Baire category theorem.

If such a space has infinitely many points, it has infinitely many isolated points. Indeed, after deleting any finite collection of isolated points, the remaining set is closed, complete, countable, and nonempty. It therefore has a point isolated in the remaining space; because the deleted set is finite, that point is also isolated in the original space.

Suppose continuous functions $f_n:\mathbb{R}\to \mathbb{R}$ converge pointwise to $f$. For each $ϵ>0$, some nonempty open interval $I$ and integer $N$ satisfy $|f_n(x)-f(x)|\le ϵ$ for every $x\in I$ and $n\ge N$. Indeed the closed sets $Q_N={\bigcap }_{n,m\ge N}\{|f_n-f_m|\le ϵ\}$ cover $\mathbb{R}$. The Baire category theorem gives an interval inside one $Q_N$, and passing $m\to \infty$ proves the assertion. The interval may depend on $ϵ$; this does not assert uniform convergence on one fixed interval.

Under the Continuum hypothesis, enumerate the meagre $F_{\sigma }$ covers by ${\omega }_1$. At stage $\alpha$ choose a new real outside all earlier covers and the previous choices. The excluded union is meagre, so the Baire category theorem makes the choice possible. Each cover then contains only countably many selected points. Arbitrary meagre sets need not themselves have only continuum many possible choices; the coded covers are what is enumerated.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

A continuous surjective linear map between Fréchet spaces is an open map. The Baire category theorem first shows that the closure of the image of every convex balanced zero-neighbourhood is a zero-neighbourhood: the target is the countable union of scalar multiples of this image, and an interior point of its closure can be translated to zero using convexity and symmetry.

To remove the closure, fix a domain zero-neighbourhood $U$ and choose convex balanced zero-neighbourhoods $U_n$ so that every series ${\sum }_n{u}_n$, $u_n\in U_n$, converges to a point of $U$. With an increasing defining sequence of seminorms $p_j$, one can ensure $p_j(u_n)<2^{-n}$ for $j\le n$ and make the finitely many seminorms defining $U$ summable within its bounds. Completeness gives convergence. Choose target zero-neighbourhoods $V_n\subset \overline{A(U_n)}$ shrinking to zero. For $y\in V_1$, choose $u_n\in U_n$ successively so that $y-A{\sum }_{j\le n}{u}_j\in V_{n+1}$. Density permits each correction. Continuity then gives $y=A{\sum }_n{u}_n\in A(U)$, proving openness.

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets. The Uniform boundedness principle says that if $V$ is a Banach space, $W$ a normed vector space, and $\mathcal{T}\subseteq B(V,W)$ is pointwise bounded, meaning ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in V$, then ${\sup }_{T\in \mathcal{T}}\Vert T\Vert <\infty$.

To prove it, let $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Each set is closed by continuity of the individual operators, and their union is $V$. The Baire category theorem gives a ball $B(x_0,r)$ inside some $E_N$. For $\Vert h\Vert <r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\Vert Th\Vert \le \Vert T(x_0+h)\Vert +\Vert T{x}_0\Vert \le 2N$ for every $T$. Scaling gives $\Vert T\Vert \le 2N/r$ uniformly. The zero-space and empty-family cases are immediate.

For the first family on a Hilbert space, fix $x$. The functionals $y\mapsto ⟨y,Tx⟩$ have operator norm $\Vert Tx\Vert$ by the Cauchy-Schwarz inequality and equality on the direction of $Tx$. Their values are pointwise bounded by the hypothesis, after taking complex conjugates if necessary. Apply the Uniform boundedness principle to obtain ${\sup }_{T\in \mathcal{E}}\Vert Tx\Vert <\infty$. Apply it a second time to the operator family itself to conclude ${\sup }_{T\in \mathcal{E}}\Vert T\Vert <\infty$.

For the diagonal hypothesis, the answer depends on the scalar field. Over $\mathbb{C}$, let $q_T(x)=⟨Tx,x⟩$, using the convention that the inner product is linear in its first variable. The complex polarization identity givesThe assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus the complex-space answer is yes, even without assuming self-adjointness.

Over $\mathbb{R}$, take $X={\mathbb{R}}^2$ andEvery diagonal value $⟨nJx,x⟩$ is zero, whereas $\Vert nJ\Vert =n$. Thus the real-space answer is no in general. Real diagonal values do not detect the skew-symmetric part of an operator. This is diagonal boundedness of operator families over real and complex Hilbert spaces.

Compactifications. A Hausdorff compactification of $X$ is a compact Hausdorff space $K$ together with a topological embedding $j:X\to K$ having dense image. We identify $X$ with that image.

For noncompact locally compact Hausdorff $X$, adjoin $\infty$ and give $X^{+}=X\cup \{\infty \}$ its one-point compactification topology: retain the open sets of $X$, and take $\{\infty \}\cup (X\setminus C)$, with $C$ compact, as a neighborhood base at $\infty$. Compact subsets are closed, making this a topology with the original subspace topology on $X$.

An open cover contains a neighborhood of $\infty$, leaving a compact subset of $X$ to cover by finitely many more members. Hence $X^{+}$ is compact. Points of $X$ are separated using its Hausdorff property; to separate $x\in X$ from $\infty$, use local compactness to choose an open $U\ni x$ with compact closure and take $X^{+}\setminus \overline{U}$ at $\infty$. Finally every neighborhood of $\infty$ meets $X$, because $X$ itself is not compact. Thus $X$ is dense and $X^{+}$ is a Hausdorff compactification with one-point remainder.

Comparison maps. The Stone-Čech compactification of a Tychonoff space has the universal property that every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Applying it to any Hausdorff compactification $K$ gives a continuous surjection $\pi :\beta X\to K$ fixing $X$. Surjectivity follows because its compact image is closed and contains the dense set $X$.

We need the fibers of compactification comparison maps propertySuppose $\pi (z)=x\in X$ with $z\ne x$. By regularity of compact Hausdorff $\beta X$, choose an open neighborhood $U$ of $x$ with $z\notin \overline{U}$. The embedding $X\subseteq K$ supplies open $V\subseteq K$ containing $x$ and satisfying $V\cap X\subseteq U\cap X$. The open set ${\pi }^{-1}(V)$ contains $z$. Density of $X$ implies $z\in \overline{{\pi }^{-1}(V)\cap X}=\overline{V\cap X}\subseteq \overline{U}$, a contradiction. Thus the displayed identity holds.

Equivalence of (a), (b), (c). Clearly (b) implies (c). If (c) holds in $K$, inverse images under $\pi$ and ${\pi }^{-1}(X)=X$ show that $X$ is a G-delta set in $\beta X$, giving (a). Conversely assume (a) and write $X={\bigcap }_n{U}_n$ in $\beta X$. The sets $F_n=\beta X\setminus U_n$ are compact and miss $X$. Their images $\pi (F_n)$ are closed in $K$ and miss $X$ by the fiber identity. Since $\pi$ is onto, they exhaust $K\setminus X$. HenceThis proves (b), for every $K$. This G-delta compactification invariance defines a Čech-complete space.

The Baire property. Suppose $X={\bigcap }_n{G}_n$ in a compact Hausdorff $K$, with $X$ dense and all $G_n$ open. Each $G_n$ is dense. For dense open subsets $U_n$ of $X$, choose open $O_n\subseteq K$ with $U_n=X\cap O_n$. Every nonempty open subset of $K$ meets $X$ in a nonempty relatively open set and hence meets $U_n$, so each $O_n$ is dense in $K$.

The assumed Baire category theorem in $K$ makes ${\bigcap }_n(G_n\cap O_n)$ dense in $K$. This set lies in $X$ and equals ${\bigcap }_n{U}_n$, so it is also dense in $X$. Therefore Čech-complete spaces are Baire spaces.

Enumerate $\mathbb{Q}=\{q_n:n\in \omega \}$. Each $\mathbb{Q}\setminus \{q_n\}$ is open and dense in $\mathbb{Q}$, while their intersection is empty. Thus $\mathbb{Q}$ is not a Baire space, and consequentlyIf $\mathbb{Q}$ were a G-delta set in $\mathbb{R}$, it would also be one in ${\mathbb{R}}^{+}$: the subspace $\mathbb{R}$ is open in its one-point compactification, and adjoining that open condition to a countable intersection preserves the G-delta set property. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and hence in ${\mathbb{R}}^{+}$, this would be a Hausdorff compactification of $\mathbb{Q}$ contradicting the preceding conclusion. Therefore $\boxed{\mathbb{Q}\text{ is not }{G}_{\delta }\text{ in }\mathbb{R}}$.

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).