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Let $X$ be a topological space and consider the set of all $k$ element multisets consisting of points from $X$. Does the topology on $X$ induce a natural topology on this space of multisets?

For example, using $\langle \dots \rangle$ to denote a multiset with specified elements, I would like to be able to say things like, $$\langle 1/n, 0, 1, 2\rangle \rightarrow \langle 0,0,1,2\rangle\quad \text{as } n \rightarrow \infty,$$

and have it be topologically meaningful.

What are the properties of such a topology on multisets? If this has been studied before, what are some good references?

This question is motivated by the following questions here and on mathoverflow about topologies on power sets of a topological space. The difference is that in this question I'm interested in multisets rather than sets, and only multisets with a fixed cardinality, rather than all multisets.

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Nick Alger
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1 Answers1

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Here is an idea. First, consider the Cartesian product $X^k$ endowed with the product topology.

It is not appropriate to identify elements of $X^k$ with the $k$-element multisets constructed from the elements of $X$, since if, for example, $k=3$, then $\langle a,a,b\rangle$ and $\langle a,b,a\rangle$ are the same multiset, whereas $(a,a,b)\neq(a,b,a)$ as elements of $X^k$ ($a,b\in X$, $a\neq b$).

However, this problem can be easily overcome by defining an equivalence relation on the elements of $X^k$ in such a way that certain distinct elements of $X^k$ are identified to conform to the structure of multisets.

Letting this equivalence relation be denoted as $\sim$, you can then try defining the quotient topology $X^k/\sim$ on the set of equivalence classes—that is, the set of $k$-element multisets.

The exact definition of $\sim$ can be formulated as follows. Suppose that $x_1,\ldots,x_k,y_1,\ldots,y_k\in X$. Let $x\equiv(x_1,\ldots,x_k)$ and $y\equiv (y_1,\ldots,y_k)$. Then, we say that $x\sim y$ if there exists a permutation (a bijective function) $\pi:\{1,\ldots,k\}\to\{1,\ldots,k\}$ such that $$y_i=x_{\pi(i)}\quad\text{for each $i\in\{1,\ldots,k\}$}.$$

triple_sec
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  • Ahh interesting. So then, for example, in the case of $(\mathbb{R}^+)^K$, if you quotient out by permutations, the result could be identified with the angular cone consisting of points where the coordinates are in sorted order – Nick Alger Oct 09 '16 at 07:03
  • @NickAlger I have not actually done the math (no pun intended), but I have a strong hunch that this quotient topology has the following property: let ${\langle x_1^{\alpha},\ldots,x_k^{\alpha}\rangle}{\alpha\in A}$ be a [net](https://en.wikipedia.org/wiki/Net(mathematics)) (a generalized sequence) of multisets. Then, this net converges to the multiset $\langle x_1,\ldots,x_k\rangle$ in the constructed quotient topology if and only if the net ${x_i^{\alpha}}_{\alpha\in A}$ converges to $x_i$ in the original topology for every $i\in{1,\ldots,k}$. – triple_sec Oct 10 '16 at 04:26