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Let $X$ be a Banach space and $B_n \subseteq B_{n+1}$ be closed sets (with non-empty interior) sets such that $$ X = \bigcup_{n=1}^\infty B_n $$ Is it true that $\{ x\in X: \|x\|=1 \} \subseteq B_N$ for some $N\in \mathbb N$?

Edit 1. This is to prove Principle of Uniform Boundedness using Baire Category Theorem. I have simplified it as above. The $B_n$ are constructed as $$ B_n = \{ x \in X: \forall A \in \mathscr A, \|Ax\| \le n \} $$ where $\mathscr A \in B(X, Y)$. Without loss of generality each $B_n$ can be taken as having non-empty interior.

My thoughts. Suppose $\epsilon$ neighborhood $V_{\epsilon}(x_0) \subseteq B_1$ then $V_{\epsilon}(x_0) - x_0 = V_\epsilon(0) \subseteq B_1 - x_0$. Scaling should give

$$ V_2(0) \subseteq B_{N \ge \frac 2 \epsilon} - x_0 $$ I don't know how translate back and include unit disc though :(

Edit 2. Seems that it is not true in general. Linearly of $A$ is very important in scaling.

Hash Nuke
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  • This looks similar to this one https://math.stackexchange.com/questions/2582687/closed-subset-of-banach-space-containing-unit-ball?rq=1 – Hash Nuke Mar 05 '23 at 22:09
  • @coudy Let $X$ is Banach and $Y$ be normed, $\mathcal A \subseteq \mathcal B(X, Y)$ s.t. $\forall x \in A, \sup { |Ax| : A \in \mathcal A} < \infty$. This comes from Principle of Uniform Boundedness – Hash Nuke Mar 05 '23 at 22:56
  • $\mathscr A \in B(X, Y)$ or $\mathscr A \subset B(X, Y)$? – geetha290krm Mar 06 '23 at 09:48

2 Answers2

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If the dimension of $X$ is zero or one, the assertion is trivially true. The following is a simple counterexample in two dimensions. Let $X = \mathbb R^2$ and $$ B_n := \left\{ (x,y) \in \mathbb R^2 \;\mid\; y \le 0 \text{ or } y \ge \frac xn \right\}. $$

I think that one should add the requirement that $B_n$ is convex, then the statement becomes true.

gerw
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Say for example that $X=\ell^2(\mathbb N)$, and $U_i=\{x\in X: \|x-e_i/3\|<1/3\}$, $i\in\mathbb N$, where $$ e_i=(0,0,\ldots,0,1,0,\ldots), $$ with the $1$ in the $i-$place. Clearly, $U_i$ is a subset of the unit ball. Set $$ U=\bigcup_{i\in\mathbb N}U_i, \quad B_n=(X\setminus U)\cup \bigcup_{1\le i\le n}\overline{U_i}. $$ Then $B_n\subset B_{n+1}$, $\bigcup B_i=X$ and $B_n$ does not contain the unit ball.

Yiorgos S. Smyrlis
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