Let $V$ be a topological vector space and $B \subseteq V$ bounded. Then the closure $\overline{B}$ is bounded.

This appears on the Wikipedia page http://en.wikipedia.org/wiki/Bounded_set_(topological_vector_space) but I am not sure how to prove it. Similar questions that I've found such as Show that the closure of a subset is bounded if the subset is bounded and Closure of a bounded set don't seem to help me because they use a different definition of boundedness and the problem seems to be easier with that definition.

My definition is: $B$ is bounded if and only if, for any open set $U \subseteq V$ with $0 \in U$, there is a scalar $\lambda$ such that $B \subseteq \lambda U$.

  • $\begingroup$ The closure of a subset include those points such that every deleted neighborhood of such a point has non-empty intersection with the original subset. Can $B$ be bounded and have a limit point arbitrarily far away from the origin? $\endgroup$ – user123641 Jun 18 '14 at 3:58
  • $\begingroup$ @Bryan It seems intuitive, but I'm uncomfortable with "arbitrarily far away from the origin" without having a metric on $V$. I'm not sure where to go with this. $\endgroup$ – user157718 Jun 18 '14 at 4:05
  • $\begingroup$ Is there some restriction on your underlying field? Is it real/complex or just any topological field? $\endgroup$ – user123641 Jun 18 '14 at 4:38
  • $\begingroup$ @Bryan Real/complex would be OK although I think that this is supposed to hold on any field with valuation $\endgroup$ – user157718 Jun 18 '14 at 4:43
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    $\begingroup$ Do you follow Rudin in assuming $V$ is T1? $\endgroup$ – Eric Towers Jun 18 '14 at 5:57

Here's a link where the result you want is prove. http://ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf Proposition 3.8-e on page 9 of the PDF (labeled page is 189.) Everything looks straightforward, but several small results are required. The main result is this: If $V$ is a neighborhood of $0$, then there exists another neighborhood $W$ of $0$ such that $W\subseteq\overline{W}\subseteq V$. That does it because $B\subseteq\lambda W\subseteq \lambda \overline{W}$ implies $\overline{B}\subseteq\overline{\lambda W}=\lambda\overline{W}\subseteq \lambda V$, which proves $\overline{B}$ is bounded.

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