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What are some good examples of separately continuous functions $f: X \times Y \rightarrow Z$ that are discontinuous at every point?

Here's a theorem to rule out some spaces: link for a reference

Theorem: Let $X$ be locally compact or completely metrizable, $Y$ compact Hausdorff, $Z$ a metric space. If $f: X \times Y \rightarrow Z$ is separately continuous, then there exists a dense $G_\delta$ subset $A$ of $X$ such that $f$ is continuous on $A \times Y$.

So no example exists for $X$, $Y$ and $Z$ satisfying the assumptions of the theorem.

spin
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    By separately continuous do you mean that when fixing $x$, the function $y\mapsto f(x,y)$ is continuous from $Y$ to $Z$, and similarly for fixing $y$? – Asaf Karagila Apr 13 '13 at 11:07
  • @AsafKaragila: Yes. – spin Apr 13 '13 at 11:09
  • What kind of spaces do you want to consider? No such $f$ exists if $X,Y,Z$ are metrizable. – Martin Apr 13 '13 at 12:20
  • @Martin: Are you sure? It seems, there is no such $f$ provided some spaces have some "complete" properties. – Alex Ravsky Apr 13 '13 at 12:29
  • @Alex: Yes, you're right. I should have said completely metrizable ($X$ and $Y$ complete should be enough). There always is a comeager set on which $f$ is continuous, but that might be empty without assuming some completeness. – Martin Apr 13 '13 at 12:39
  • @Martin: Any topological spaces. See my edit – spin Apr 13 '13 at 12:44
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    There should be a lot of such examples in the papers by V. K. Maslyuchenko and his son, O. V. Maslyuchenko. They both are powerful mathematicians and have a family “tradition” or, even, a school, devoted to different weak continuity of maps on the products of different topological spaces. :-) So I suggest you to search and to look their papers devoted to separately continuous not (joint) continuous maps. – Alex Ravsky Apr 13 '13 at 12:43
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    @spin: If any spaces can satisfy you, you may take $X=Y=Z=\mathbb{R}$ in the cofinite topology (that is, a nonempty set $A\subset\mathbb{R}$ is open iff $\mathbb{R}\backslash A$ is finite) and $f(x,y)=x+y$ for each $x,y\in\mathbb{R}$. :-) – Alex Ravsky Apr 14 '13 at 15:19
  • @AlexRavsky: Thanks, that's a very simple example. I believe it should work even when we replace $\mathbb{R}$ by any infinite abelian group. – spin Apr 14 '13 at 16:36
  • It seems so for any infinite group $G$, since for each cofinite subsets $U,V$ of $G$ we have that $UV=G$, therefore the multiplication on $G\times G$ maps each nonempty open set onto the whole $G$. – Alex Ravsky Apr 14 '13 at 16:50

1 Answers1

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Answering so that this isn't unanswered. In the comments, Alex Ravsky suggested the following example which was just what I needed.

Consider any infinite group $G$ equipped with the cofinite topology. Then the multiplication map $G \times G \rightarrow G$ defined by $(x,y) \mapsto xy$ is separately continuous everywhere, but is not jointly continuous at any point.

spin
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