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
$$
\mathrm{dist}([s\_1,s\_2,\ldots],[t\_1,t\_2,\ldots]) := \inf\_{\pi} \sum\_i d(s\_i,t\_{\pi(i)}),
$$
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. 



The Metrizability of Symmetric Products of Metric Spaces
The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.


Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x\_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x\_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x\_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x\_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x\_n)$ converges to $[e]$. More generally, this argument shows that if a $T\_1$ space $A$ is a colimit of subspaces $A\_0\subset A\_1\subset \dots$ such that the interiors of the $A\_n$ do not cover $A$, then $A$ cannot be first countable.


As a final note, let $SP\_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP\_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $(X,e)$ is a pointed metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi\_n(SP\_d(X,e))=\pi\_n(SP(X,e))=H\_n(X,e)$.


