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In a metrizable topological vector space X with the metric d, a subset A is said to be bounded if it can be absorbed by any neighbourhood of 0 and a subset A is said to be d-bounded if its diameter with respect to the metric d is finite. Boundedness always implies d-boundedness, but the converse is not true.

I am looking for a condition for which d-boundedness implies boundedness. In the Wikipedia, in the section "Topological vector spaces'', there is a statement, "The two notions of boundedness coincide for locally convex spaces''. But there is no reference for it there. Can somebody give some reference or some hint to prove this statement?

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closed as off-topic by YCor, Joseph Van Name, abx, Sean Lawton, Pace Nielsen Feb 20 at 16:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Joseph Van Name, abx
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The Wikipedia quote you mention is about boundedness for linear operators, not for sets. Your statement is false if $X$ is not normed, see Bourbaki's Topological Vector Spaces, ch. III, §1, Remark 1. $\endgroup$ – abx Feb 13 at 5:04
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For boundedness of sets the statement is false. The Wikipedia quote is for linear operators.

A counterexample for sets: $X=\mathbb{R}^\omega$ in the product topology is a metric locally convex TVS. No neighbourhood of $0$ (like the open balls which are $d$-bounded) can be "absorbing-bounded" (Because it contains a product basic open set which has almost all factors equal to $\mathbb{R}$), so $X$ is not normable.

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