Compactness of symmetric power of a compact space

Question

Suppose I have a compact metric space $(X,d)$ and let $\mathcal{X}=X^K$ be the product space. Consider the equivalence relation $\sim$ on $\mathcal{X}$ given as: for $\alpha,\beta\in \mathcal{X}$, $\alpha\sim\beta$ iff there exists a permutation $\tau$ of $\{1,\dots,K\}$ such that $\alpha_i = \beta_{\tau(i)}$ for all $i$. Consider the quotient space $\mathcal{X}/\sim$ and define a metric $\rho$ on this quotient space as $\rho([\alpha],[\beta]) = \min_{\tau} \sum_i d(\alpha_i, \beta_{\tau(i)})$.

I could show this defines a well-defined valid metric. I also know that product of compact spaces under the product metric is compact (Tychonoff's theorem) and for a compact topological space $Q$ with an equivalence relation $R$, the quotient space $Q/R$ is also compact. But are there any results showing whether $(\mathcal{X}/\sim, \rho)$ is compact under the topology induced by this metric (necessary and/or sufficient conditions if any)?

Answer 1

Here's a direct proof (although showing your metric yields the topology of $\mathcal X/\sim$ is more useful). In a metric space, compactness is equivalent to sequential compactness: every infinite sequence of points has a convergent subsequence. Let $x_n\in X^K$ correspond to an arbitrary sequence $[x_n]\in\mathcal X/\sim$. Since $X^K$ is compact, let $y_n$ be a subsequence converging to $x\in X^K$.

I claim $[y_n]$ converges to $[x]$. To see this, let $\epsilon>0$, and choose $N$ with $d(y_n(i),x(i))<\epsilon/K$ for each $1\leq i\leq K$ and $n\geq N$. Then $\rho([y_n],[x])\leq\sum_i d(y_n(i),x(i))<\epsilon$ for all $n\geq N$.

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This also basically shows that the map $x\mapsto[x]$ is continuous, which is another angle, since continuous images of compact spaces are compact.