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Let $(X, \tau)$ be a topological vector space. We say that a set $E \subset X$ is bounded if, for all neighborhood $V$ of $0$, there exists $r > 0$ such that $E \subset rV$. Now let $\{x_n\}$ be a Cauchy sequence on $X$, that is, given an neighborhood V of $0$, there exists an $N_0 \in \mathbb{N}$ such that $x_n - x_m \in V$, for all $n,m \geq N_0$. I have to proof that $E = \{x_n : n \in \mathbb{N}\}$ is a bounded set. I'm seeing the proof on Rudin, Functional Analysis, 1973, on page 22, and I got stuck on a step.

First he takes $V$ and $W$ two balanced neighborhoods of $0$ with $V + V \subset W$ (for balanced we mean, $\lambda y \in V$, $\forall y \in V$ and $|\lambda| \leq 1$). Then, there exists an $N \in \mathbb{N}$ such that $x_n - x_m \in V$, for all $n,m \geq N$ (this is the definition of a Cauchy sequence on a topological vector space). In particular, $x_n \in x_N + V$, for all $n \geq N$. Here comes my doubt: He takes $s > 1$ such that $x_N \in sV$, that is, $\frac{1}{s} x_N \in V$. I could't understant this. If $x_N$ was in $V$, this would follow from the fact that $V$ is balanced, but I think this is not the case.

Any help with this proof, or another reference with this would be nice.

ThiagoGM
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1 Answers1

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Note that $V$ is an open set containing $0$ and that $0 \cdot x_N = 0$. The map $\tau: \lambda \mapsto \lambda \cdot x_N$ is continuous from $\mathbb{R}$ to $X$ so that $\tau^{-1}(V)$ is an open set in $\mathbb{R}$ containing $0$.

This tells you that there exists $\delta > 0$ such that if $|\lambda| < \delta$ then $\lambda \cdot x_N \in V$.

Rhys Steele
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  • Dear Rhys Steele, thank you very much, this helped me. – ThiagoGM Jan 11 '23 at 17:15
  • Is there an easy way to see that $\tau$ is continous? I see that $\tau(\lambda) = M_{\lambda}(x_0)$, and I know that $M_{\lambda} : X \rightarrow X$ is continous. But I don't know how to proceed. – ThiagoGM Jan 11 '23 at 17:33
  • Perhaps what you are missing is that the definition of a topological vector space tells you the stronger fact that $M: \mathbb{R} \times X \to X$ defined by $M(\lambda, x) = M_\lambda(x)$ (using your notation) is continuous? $\tau$ is then just a section of this map (where we fix $x_N$). – Rhys Steele Jan 11 '23 at 17:44
  • Once again, thank you. I just forgot that the application $M$ is defined on $\mathbb{R} \times X$. – ThiagoGM Jan 11 '23 at 17:47
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    I have a question. Why we should choose balanced neighborhoods $W,V$when we prove that Cauchy sequence is bounded. Why it must be balanced? – xyz Jun 23 '23 at 04:00