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I have read that if the subsets $A$ and $B$ of a topological vector space are bounded, i.e. for any neighbourhood $U$ of $0$ there is an $n>0$ such that, for all $|\lambda|\geq n$, $M\subset\lambda U$, then $A+B$ is bounded.

While it's so trivial when normed spaces are considered, I cannot see why it holds in any topological linear space...

$\infty$ thanks!

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Let $U$ be a neighbourhood of $0$. Choose a balanced neighbourhood $V$ of $0$ such that $V+V\subset U$. Pick $n_1,n_2 > 0$ such that $A\subset \lambda V$ for $\lvert\lambda\rvert \geqslant n_1$ and $B\subset \lambda V$ for $\lvert \lambda\rvert \geqslant n_2$. Let $n = \max\{n_1,n_2\}$. Then for $\lvert\lambda\rvert \geqslant n$ we have ...

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  • $\begingroup$ Thank you so much!!! You're using the existence of a balanced neighbourhood contained in a neighbourhood of 0 that you taught me! I see that, for such a $\lambda$, $A+B\subset\lambda V+\lambda V$, but I'm not sure how to find a $\lambda'$ such that $\lambda V+\lambda V\subset \lambda' U$, though I guess that $V$'s balancedness is the key... $\endgroup$ – Self-teaching worker Sep 4 '14 at 16:16
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    $\begingroup$ That's where the balancedness comes in. Since $V$ is balanced, we have $\lambda V + \lambda V = \lambda(V+V)$ [one easy way to prove that is to prove that $\lambda V = \lvert\lambda\rvert V$ for balanced $V$]. An alternative proof that doesn't use balanced neighbourhoods [directly] is to note that $+ \colon E\times E \to E$ is a continuous linear map, $A\times B$ is bounded in $E\times E$, and continuous linear maps map bounded sets to bounded sets. $\endgroup$ – Daniel Fischer Sep 4 '14 at 17:34

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