What is the motivation behind the definition of bounded set in a topological vector space? The definition is different from the boundedness definition in metric space.

Why is it not simply defined as follows: A set is bounded if there is a neighbourhood around zero which contains that set ?

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    $\begingroup$ Look at the column on the right: all titles begin with an uppercase letter. Capitalization is a detail, but it affects the appearance of the site. I corrected it here. $\endgroup$ – user147263 Aug 13 '14 at 4:39

In a metric space you have a way to measure distance and thus the definition of boundedness is, in a sense, obvious. In a topological vector space there are no distances, so of course the definition will be different than that given for metric spaces simply since you can't state the metric definition without a metric. However, if you think about it, the intuition is still the same. Try contemplating the definition. Stare at it for a while, translate it back to a familiar (metric!) topological vector space until you realize that it says essentially the same thing only using a more general language, since that is all you have.

  • $\begingroup$ Why is it not simply defined as follows: A set is bounded if there is a neighbourhood around zero which contains that set ? $\endgroup$ – Anonymous Aug 13 '14 at 10:37
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    $\begingroup$ Because the full set is a neighborhood of zero. $\endgroup$ – Jonas Dahlbæk Aug 13 '14 at 11:00
  • $\begingroup$ @user161825: sorry, my bad, that was trivial. But still I don't understand the reason behind such generalization. As Ittay Weiss suggested the special case trivially follows from the general case but my question was why such generalization ? $\endgroup$ – Anonymous Aug 13 '14 at 12:14
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    $\begingroup$ It is just sort of the right thing to do. For instance, a sequence $(x_n)$ is bounded if and only if for any complex sequence $\alpha_n$ with $\alpha_n\rightarrow 0$, then $\alpha_n x_n\rightarrow 0$, as you would expect of a 'bounded' sequence. Similarly, $E$ is bounded if and only if every sequence in $E$ is bounded. $\endgroup$ – Jonas Dahlbæk Aug 13 '14 at 12:28
  • $\begingroup$ @user161825: Is there any notion of boundedness in case of only topological (not vector) space ? $\endgroup$ – Anonymous Aug 13 '14 at 17:20

A bounded set $S$ in a topological vector space is one for which, given any neighborhood $N$ of the zero vector, there is a scalar $\alpha$ such that $S \subset \alpha N$. For more: http://en.wikipedia.org/wiki/Bounded_set_(topological_vector_space)

The motivation for this definition is fairly straightforward--if a "stretching" of a neighborhood of the zero vector contains the set, it is, in that sense, "bounded". In this way, the definition of boundedness is made not to depend upon distance, but solely on the notions of scaling, and of open sets.

  • $\begingroup$ I know that. I wanted to know about the motivation behind such definition as I mentioned in the question. $\endgroup$ – Anonymous Aug 13 '14 at 4:13
  • $\begingroup$ Oh, my bad. I can't into reading. $\endgroup$ – Fargle Aug 13 '14 at 4:27

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