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Let $(X,d)$ be a complete separable metric space, and endow $Iso(X,d)$ with the pointwise convergence topology.

I've seen a few sources say this is clearly a Polish group, but why is this this the case?

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    $\begingroup$ This is sort of an exercise, I think. One approach is to fix a countable dense subset $E \subset X$ and show that the "restriction to $E$" map gives an embedding of $\mathrm{Iso}(X,d)$ into the Polish space $X^E$ (with its product topology) whose image is closed (hence complete). A useful lemma is to show that a sequence of isometries $f_n$ converging pointwise on $E$ must actually converge pointwise on $X$, and the limit is again an isometry. $\endgroup$
    – Nate Eldredge
    Commented Oct 24, 2016 at 3:07
  • $\begingroup$ Indeed, by an Arzela-Ascoli type argument, the convergence is uniform on compact sets. This is helpful in showing that the operation of composition is jointly continuous. $\endgroup$
    – Nate Eldredge
    Commented Oct 24, 2016 at 3:08

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