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

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?

  • $\begingroup$ 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

$X$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.