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?


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

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.