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The continuum hypothesis holds. Let $A$ be the assumed Lusin set of cardinality $\mathfrak{c}=2^{{\aleph }_0}$. If $\mathfrak{c}>{\aleph }_1$, choose $B\subseteq A$ with $|B|={\aleph }_1$. The size assumption makes $B$ a meagre set, but then $A\cap B=B$ is uncountable, contradicting the Lusin set property. ThusOne can also use (d) directly: well-order $A$ in type $\mathfrak{c}$. Each proper initial segment has size below $\mathfrak{c}$, so is meagre and consequently countable, since it is a subset of the Lusin set. These initial segments form a linear increasing union of countable sets equal to $A$. The bound in (d) gives $\mathfrak{c}\le {\aleph }_1$.

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

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.

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.