I'm reading about Topological Vector Spaces (using Treves's book) and I failed to answer the question on title. I'll be more precise:

1) We say that a TVS $E$ is complete if every Cauchy filter in $E$ is convergent.

2) Given $E_1$ and $E_2$ two TVS, we consider $E=E_1 \times E_2$ with the product topology. With this topology and the natural structure of vector space, $E$ is a TVS.

In the Bourbaki's book is proved that it's true when we use the notion of uniform structure. But looking at page 140, ex. 5 of Topological Vector Spaces and Distributions (by John Horvath, where he uses just the definition 1), he states just the converse.

So, is it true that $E$ is complete when $E_1$ and $E_2$ are complete? If it's not true, are there some example?

  • $\begingroup$ The formulation "he states just the converse" confused me a bit. At first I thought that you mean that the author states that this is not true. But, when I looked there, I see that they only give there the converse implication: if $E_1\times E_2$ is complete then the spaces $E_{1,2}$ are complete. (Although it is stated there more generally - for arbitrary system of spaces, not only finitely many of them.) $\endgroup$ – Martin Sleziak Aug 13 '15 at 14:17
  • $\begingroup$ You're right. The if the product of an arbitrary family is complete, then each 'component' of the product is complete. I was trying to prove the other direction, but even when the family is indexed by 2 elements I've failed. The 'uniform structure' mentioned in Bourbaki made me thought that this direction is false, but I havent found an example. $\endgroup$ – Grotrev Aug 13 '15 at 14:23
  • $\begingroup$ The Horvath's book you quoted says on p.139: "It follows that if for each $i\in I$ the set $A_i$ is a complete subset of $E_i$, then the subset $A=\prod_{i\in I} A_i$ of $E$ is complete." So the reason that the exercise only mentions one implication might be that the converse implication has already been proved. $\endgroup$ – Martin Sleziak Aug 13 '15 at 15:57
  • $\begingroup$ I didn't see that. Thank you! $\endgroup$ – Grotrev Aug 13 '15 at 18:39

Yes the product of tow complete topological vector space is complete. The very simple example is considering $\mathbb R $ and $\mathbb R^2$

  • $\begingroup$ Why?why?why?why? $\endgroup$ – Darman Nov 16 at 10:23

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