I want to prove that a product $\prod_{i\in I}X_i$ of topological vector spaces is pseudonormable only if a finite number of the factor spaces are also pseudonormable and the rest have the trivial topology. Could anyone point me in the right direction?

  • $\begingroup$ close to your question math.stackexchange.com/questions/139168/… $\endgroup$ – Norbert Dec 27 '12 at 23:02
  • $\begingroup$ Pardon my ignorance, what is a pseudonormable space (is it the same as a seminormable space or is it something else)? Google is not of much help: most results are this question or links to it and I managed to find two or three articles which do contain the word but no definition. $\endgroup$ – Martin Dec 27 '12 at 23:32
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    $\begingroup$ It's just like a normal norm but can be zero at nonzero x. So it induces non-hausdorff topologies. $\endgroup$ – Parakee Dec 27 '12 at 23:39
  • $\begingroup$ Alright. Thanks! $\endgroup$ – Martin Dec 27 '12 at 23:43

I guess this is a simple combination of the following well-known facts:

  1. A TVS $X$ is seminormable (or, as called above, pseudonormable) iff it is locally convex and has a bounded $0$-neighbourhood (i.e. absorbed by every other $0$-neighbourhood).

  2. A set $B$ is bounded in a product space $\prod_i X_i$ with product topology iff $B=\prod_i B_i$ where each $B_i$ is bounded in $X_i$

  3. If a TVS $X$ is bounded, its topology is the trivial topology (i.e. $\emptyset$ and $X$ are the only open sets).

So if $\prod_i X_i$ is seminormable, then there is a $0$-neighbourhood $B=\prod_i B_i$ where each $B_i$ is bounded in $X_i$ and hence for all but at most finitely many $i$ you have $B_i = X_i$.


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