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I have been looking at Topological Groups, and I recently read about the group $\operatorname{Homeo}(X)$ of all homeomorphisms of $X$ onto itself. In particular, when $X$ is a metric space. The subgroup $\operatorname{Is}(X)$ of all isometries of $X$ onto itself equipped it with the topology of pointwise convergence turns out to be a topological group. I verified that the multiplication and inverse operations are continuous.

I've read that if $X$ is a normal space, then $\operatorname{Homeo}(X)$ equipped with the closed-base topology is a topological group. Also, if $X$ is a compact Hausdorff space, then $\operatorname{Homeo}(X)$ equipped with the compact-open topology is a topological group.

My question is about what happens if we equip $\operatorname{Homeo}(X)$ (where $X$ is a metric space) with the topology of pointwise convergence. Is this still a topological group? Is there a specific space $X$ such that $\operatorname{Homeo}(X)$ equipped with the topology of pointwise convergence is not a topological group?

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up vote 2 down vote accepted

If $X=R^2$, then $Homeo(X)$ with the topology of pointwise convergence is not a topological group. In fact, the multiplication is not jointly continuous.

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And neither is the inverse, IIRC –  Henno Brandsma Mar 11 '12 at 21:25
    
I was able to find a counter example to construct a function that is not jointly continuous on the the plane. Thanks for the suggestion. –  josh Mar 13 '12 at 16:02
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