Timeline for A topological vector space $X$ is separable if its dual space $X^*$ is separable?
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| Apr 1, 2020 at 1:34 | comment | added | Robert Furber | @მამუკაჯიბლაძე In terms of counterexamples, in Schaefer it is just not stated that the two are the same, and the fact that they are the same for a $\aleph_0$-fold coproduct is set as an exercise. But in this answer and this comment by Paul Garrett links are provided to counterexamples for the case of an uncountable coproduct. | |
| Mar 31, 2020 at 17:09 | comment | added | მამუკა ჯიბლაძე | Thank you for the explanation. It is subtler than I expected, I thought you meant that the topological vector space coproduct is locally convex in this case, which is presumably not true. | |
| Mar 31, 2020 at 16:31 | comment | added | Robert Furber | @მამუკაჯიბლაძე That depends on what you mean by "initial topology". It is true that every linear functional is continuous (and it has to be, to be a coproduct). However, the topological vector space initial topology and locally convex initial topology are not identical for uncountable coproducts. This is why I specified the "locally convex coproduct topology" (which should be considered as one word, rather than as saying that the copduct topology is locally convex). See Schaefer's Topological Vector Spaces section II.6. The coproduct is called the "locally convex direct sum" there. | |
| Mar 31, 2020 at 16:19 | comment | added | მამუკა ჯიბლაძე | For $X^*$ to be ${\mathbb R}^{\mathbb R}$, every linear functional on $X$ must be continuous. That is, the initial topology must be locally convex. Is this obvious? | |
| Mar 31, 2020 at 15:50 | comment | added | Nik Weaver | Very cool example! | |
| Mar 31, 2020 at 15:35 | history | answered | Robert Furber | CC BY-SA 4.0 |