# Two notions of boundedness in metrizable topological vector space.

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 wiki, 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?

• Do you want to see it in it the most general case, or is the Boundlessness between Banach spaces enough? – Keen-ameteur Feb 12 at 11:27
• In the general case. – Infinite Feb 12 at 13:49
• What two notions of boundedness are supposed to be equivalent? metric and linear space? That's false. – Henno Brandsma Feb 12 at 15:29
• Boundedness and d-boundedness for a subset in a metrizable topological vector space. I edited my question. – Infinite Feb 12 at 15:50
• @Infinity $\mathbb{R}^\omega$ is a counterexample in the usual metric. A locally convex space with a bounded neighbourhood is normable which this space isn’t. – Henno Brandsma Feb 12 at 18:19

The simplest example is $$\mathbb{R}$$ with the metrics $$d_1(x,y)=|x-y|$$ and $$d_2(x,y)=|\arctan(x)-\arctan(y)|$$. These two metrics define the same topology. The definition of bounded with absorption is a topological property and does not depend on which metric you choose. $$d$$-boundedness depends on the metric.