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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.

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.

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.

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.

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.