The (infinite) symmetric product of a based topological space $(X,e)$, denoted by $\mathrm{SP}(X,e)$, can be viewed as the topological space of ''multisets'' in $X$ containing the base point $e$ infinitely many times (please see http://en.wikipedia.org/wiki/Infinite_symmetric_product for the precise definition). The following is my question:
Provided that $X$ is metrizable, can we say that $\mathrm{SP}(X,e)$ is metrizable in general?
Due to my poor knowledge I have absolutely no idea as to how the symmetric product of simple spaces look like. However, I suppose I can explain how I come up with this question at least. Let $(X,d)$ be a metric space with a base point $e$. An arbitrary element $S$ in $\mathrm{SP}(X,e)$ admits a representation $S = [s_1,s_2,\ldots]$, where $(s_i)_{i \in \mathbb{N}}$ is a sequence in $X$ with the property that all but finitely many terms of the sequence are the base point $e$. We can now define a metric on $\mathrm{SP}(X,e)$ by $$ \mathrm{dist}([s_1,s_2,\ldots],[t_1,t_2,\ldots]) := \inf_{\pi} \sum_i d(s_i,t_i), $$$$ \mathrm{dist}([s_1,s_2,\ldots],[t_1,t_2,\ldots]) := \inf_{\pi} \sum_i d(s_i,t_{\pi(i)}), $$ where the inf is taken over all permutations $\pi$. In my masters thesis (on functional analysis) I had to show that the fundamental group of $\mathrm{SP}(X,e)$, equipped with the above metric topology, is the first homology group of $X$ (provided that $X$ is both path connected and locally simply connected). I have met one algebraic-topologist who made a somewhat interesting remark that this result is analogous to the Dold-Thom theorem (please see http://en.wikipedia.org/wiki/Dold%E2%80%93Thom_theorem). This is how I started finding a relationship between the topology induced by this metric and the standard topology on $\mathrm{SP}(X,e)$. The Dold-Thom theorem seems pretty famous, and so I thought the metrizability of the symmetric product of a metric space might have been well studied. Sadly, I know nothing about topology, so this is how I ended up using this website.
Cheers.
