Is this set in $C[0,1]$ a countable union of nowhere dense sets? $\mathcal{A}=\lbrace x=x_y(t)=\int^{t^2}_0 y(\tau)d\tau : ||y||\le 1\rbrace$
space $X=C[0,1]$
$A_n$ is a nowhere dense if the interior of the closure is empty,  $\operatorname{Int}\overline{A_n}=\emptyset$,
is it possible that: $\mathcal{A}=\cup^\infty_{n=1} A_n$, countable union and $A_n$ is nowhere dense?
 A: $X$ is a complete metric space and closed subsets of complete metric spaces are complete. As a consequence, if $\mathcal A$ is closed it cannot be a countable union of nowhere dense sets by the Baire category theorem. 
Is $\mathcal A$ closed? A set is closed if and only if it contains all its limit points. Let $x_n \in \mathcal A$ be a sequence converging to some $x \in X$. The point $x$ is in $\mathcal A$ if there exists $y \in X$ such that $x(t) = \int_0^{t^2} y(\tau) d\tau$ and $\|y\|_\infty \le 1$. Let $y_n$ denote the elements in $X$ such that $\|y_n\|_\infty \le 1$ and $x_n(t) = \int_0^{t^2} y_n(\tau) d\tau$. Let $y$ be the limit of $y_n$ in the $\sup$ norm. Let $\varepsilon > 0$ and $N$ be so large that $\|x_n-x\|_\infty < {\varepsilon \over 2}$ and $\|y_n-y\|_\infty < {\varepsilon \over 2}$.
Then 
$$ \begin{align}
\|x-\int_0^{t^2} y(\tau) d\tau\|_\infty &\le
 \left\| x-x_N\right\|+\left\| x_N-\int_0^{t^2} y(\tau) d\tau\right\|\\
&= \left\| x-x_N\right\|+\left\| \int_0^{t^2} y_N(\tau) d\tau-\int_0^{t^2} y(\tau) d\tau\right\| \\
&<{\varepsilon \over 2}+\left\| \int_0^{t^2} y_N(\tau) d\tau-\int_0^{t^2} y(\tau) d\tau\right\| \\
&={\varepsilon \over 2}+\left\| \int_0^{t^2} y_N(\tau)-y(\tau) d\tau\right\| \\
&<{\varepsilon \over 2} + {\varepsilon \over 2}= \varepsilon
\end{align}$$
Since $\varepsilon$ was arbitrary, $x = \int_0^{t^2} y(\tau) d\tau$. Therefore $\mathcal A$ is closed and it cannot be a countable union of nowhere dense sets.
