Let $X$ be a connected Hausdorff space. It is well-known that the $n$-fold symmetric product $\mathcal{F}_n(X) := \{A\subseteq X : 0<|A|\leq n\}$ is a connected space equipped with the Vietoris topology. ¿Has it been proven that the $n$-fold deleted symmetric product $[X]^{n} := \{A\subseteq X : |A| = n\}$ is also connected when equipped with the Vietoris topology? A colleague and I are under the impression that someone already did this, but we haven't found a reference to confirm it.
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3$\begingroup$ If I am reading your question correctly, it is a duplicate of this one math.stackexchange.com/q/148868 , which has some informative answers. $\endgroup$– Gregory AroneCommented Mar 16 at 7:49
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$\begingroup$ Maybe I'm missing something, but doesn't the following argument work? Let $U=\langle U_0,\ldots, U_n\rangle\subseteq [X]^n$ be an open set which is nonempty and not the whole space. We show it is not closed. By assumption there must be $i$ such that $U_i\neq X$, and since $X$ is connected, $U_i$ is not closed. Assume wlog $i=0$ Let $x_\lambda\to x$ be a net in $U_0$ converging to a point $x$ not in $U_0$. Let $u_j\in U_j\setminus U_0$, for $j>0$ be arbitrary. The net $E_\lambda=\{x_\lambda,u_1,\ldots,u_n\}$ converges to $\{x,u_1,\ldots,u_n\}$ which is not in $U$. $\endgroup$– Alessandro CodenottiCommented Mar 16 at 9:42
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$\begingroup$ @AlessandroCodenotti: To start with, not all open sets are of that form. $\endgroup$– Andy PutmanCommented Mar 16 at 14:34
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$\begingroup$ @GregoryArone I don't have a proof available, but I think it's not the same topological space, is it? The space $[X]^n$ and the space $C_n(X)$ are not homeomorphic. $\endgroup$– PelusoCommented Mar 16 at 15:53
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2$\begingroup$ Have a look at this question and answer mathoverflow.net/questions/293942/… . I think the answer more or less tells you that they are homeomorphic, at least when $X$ is a Hausdorff space. $\endgroup$– Gregory AroneCommented Mar 16 at 17:40
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