Natural Choice of Topology for Free Monoid on a Space

Question

Suppose $X$ is a metric space, and let $X^*$ denote the free monoid on $X$, that is the monoid consisting of all finite strings of elements of $X$, with string concatenation as the monoid operation (and with the empty word as identity).

Question: Is there a natural choice of topology on $X^*$ that reflects the topology on $X$?

Answer 1

The free monoid on a topological space $X$ is just the disjoint union

$$\bigsqcup_{n \ge 0} X^n$$

so it has a natural topology given by the disjoint union of the product topologies. Unfortunately I don't think it's useful for very much. A variation of this construction involving 1) commutativity and 2) fixing a basepoint $x \in X$ and requiring it to act as the identity produces a more interesting construction called the infinite symmetric product which, under mild hypotheses, has the property that its homotopy groups compute the reduced homology of $X$ (the Dold-Thom theorem).

Answer 2

Another possibility. Let $(X,d)$ be the metric space. Choose $x \in X$. As $X^*$ is the set of finite sequences with concatenation, it can be embedded, as a set, into $X^{\mathbb N}$, i.e., the set of infinite sequences, by $$ (x_1, \ldots, x_n) \mapsto (x_1, \ldots, x_n, x, x, \ldots). $$ Then, we can define a metric on $X^{\mathbb N}$ by $$ d'((x_1, x_2, \ldots), (y_1, y_2, \ldots)) = \sum_{i=1}^{\infty} \frac{1}{2^{i}} \frac{d(x_i, y_i)}{1 + d(x_i, y_i)}. $$ This gives an embedding of the metric space $(X, d)$ into $(X^{\mathbb N}, d')$ which yields the product topology on $X^{\mathbb N}$.