# Two notions of boundedness in metrizable topological vector space [closed]

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 NielsenFeb 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
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• 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. – abx Feb 13 at 5:04

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.