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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 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?

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  • $\begingroup$ Do you want to see it in it the most general case, or is the Boundlessness between Banach spaces enough? $\endgroup$ – Keen-ameteur Feb 12 at 11:27
  • $\begingroup$ In the general case. $\endgroup$ – Infinite Feb 12 at 13:49
  • $\begingroup$ What two notions of boundedness are supposed to be equivalent? metric and linear space? That's false. $\endgroup$ – Henno Brandsma Feb 12 at 15:29
  • $\begingroup$ Boundedness and d-boundedness for a subset in a metrizable topological vector space. I edited my question. $\endgroup$ – Infinite Feb 12 at 15:50
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Feb 12 at 18:19
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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.

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