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I'm trying to do an exercise from Rudin's "Functional analysis, 2nd edition". It is question 6 from the first chapter:

"Prove that a set E in a topological vector space is bounded if and only if every countable subset of E is bounded"

My efforts so far:

If $E$ is bounded, then if $U$ is any neighbourhood containing $0$, $E\subset tU$ for all $t$ large enough and positive. $A\subset E$, and so for any $U$ a neighbourhood of $0$, $A\subset tU$ for $t$ large enough.

I'm having trouble with the "only if" part. I cannot see where the "countability" of $A$ comes in:

If $A$ is a bounded countable subset of $E$, then $A\subset tU$ for $t$ large enough, and $U$ is as before. Now, somehow I need to argue that $A\subset E \subset tU$.

Any hints would be appreciated, thanks!

MathsStudent
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1 Answers1

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Hint. Try contradiction. If $E$ is not bounded, there is some $U$ such that $E \not\subseteq nU$ for all $n \in \mathbb N$. Choose $a_n \in E \setminus nU$. Is $A = \{a_n \mid n \in \mathbb N\}$ bounded?

martini
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  • Ah great, thanks a lot! Could you check if my argument is rigorous? So, if $E$ is not bounded, then there exists a countable set $A$ which is as you specified above. This countable set is a subset of $E$, and is not bounded, because for any $U$, a neighbourhood of $0$, there does not exist a $t$ large enough such that $A\subset tU$. – MathsStudent Jul 11 '14 at 11:44