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$\mathcal{A}=\lbrace x=x_y(t)=\int^{t^2}_0 y(\tau)d\tau : ||y||\le 1\rbrace$

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?

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  • $\begingroup$ Normally such "rare" sets (French?) are called nowhere dense. A countable union of them is called "a set of first category", or just a "first category set". So you're asking whether $\mathcal{A}$ is of first category. $\endgroup$ – Henno Brandsma Apr 23 '14 at 11:44
  • $\begingroup$ yes, indeed. thats my struggle $\endgroup$ – 104078 Apr 23 '14 at 11:48
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$X$ 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.

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  • $\begingroup$ Be careful! Every proper closed subspace $L$ of a Banach space $X$ is nowhere dense (hence of first category) in $X$ but on the other hand it is of second category in itself. $\endgroup$ – Jochen Apr 25 '14 at 7:08

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