Let $(X,\tau)$ be a Hausdorff space. Let $[X]^2 = \big\{\{x,y\}: x,y\in X \land x\neq y\big\}$. For $U,V\in \tau$ with $U\cap V = \emptyset$ we set $[U,V] = \big\{\{x,y\} \in [X]^2: x\in U\land y\in V\big\}$.

We endow $[X]^2$ with the topology $[\tau]^2$, which is generated by $\{[U,V]: U,V\in \tau\land U\cap V =\emptyset\}$.

Is there a non-discrete Hausdorff space $X$ with more than 3 points such that $X\cong [X]^2$?

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    $\begingroup$ Your notation $\{x,y\}$ instead of $(x,y)$ suggest that you are considering that $\{x,y\} = \{y,x\}$ while your definition of the topology seem to suggest that it is not the case, could you clarify this point ? $\endgroup$ – Simon Henry May 27 '15 at 9:46
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    $\begingroup$ @SimonHenry: $[X]^2$ is a very popular notation for the set of all subsets of $X$ of size $2$. I don't see how the definition of the topology suggests otherwise. In fact $[U,V]=[V,U]$. $\endgroup$ – Ramiro de la Vega May 27 '15 at 11:03
  • $\begingroup$ Thank you for clarifying. For me it suggested otherwise because if I parse the definition naively, $\{x,y\} \in [U,V]$ mean $x \in U$ and $y \in V$ while the $\{y,x\} \in [U,V]$ mean $y \in U$ and $x \in V$ and those two are clearly not equivalent, so it mean that $\{x,y\}$ and $\{y,x\}$ have to be different. Now If it is a standard notion I understand that people use this kind of notational short cut. $\endgroup$ – Simon Henry May 27 '15 at 11:14
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    $\begingroup$ Now I see the confusion; but note that $\{x,y\} \in [U,V]$ iff $x \in U \land y \in V$ or $y \in U \land x \in V$. $\endgroup$ – Ramiro de la Vega May 27 '15 at 11:20
  • $\begingroup$ To see if I have got this right: if we take $X = [0,1]$, is $[[X]]^2$ homeomorphic to the triangle $\{(x,y) \in [0,1]^2 : x < y\}$ (with its Euclidean topology)? $\endgroup$ – Nate Eldredge May 27 '15 at 15:18

It is easy to see that the operation $X \mapsto [X]^2$ preserves the properties: countable, second-countable, regular, no isolated points. Hence $\mathbb{Q} \cong [\mathbb{Q}]^2 $.

  • $\begingroup$ You are saying that $\mathbb{Q}$ is the unique space with these properties (up to homeomorphism)? Where can one find a proof of this? $\endgroup$ – Nate Eldredge May 27 '15 at 18:08
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    $\begingroup$ @Nate: Yes, this was proved by Sierpinski. See at.yorku.ca/p/a/c/a/25.pdf for a proof. $\endgroup$ – Will Brian May 27 '15 at 20:12

Write $F(X)=[X]^2$. Then $F$ is an endofunctor on the category of Hausdorff spaces and inclusions of clopen subspaces, and it preserves filtered colimits. Given any Hausdorff space $X$ with an inclusion of $X$ as a clopen subspace of $F(X)$, it follows (as in this answer) that $F^\omega(X)=\operatorname{colim} F^n(X)$ satisfies $F(F^\omega(X))\cong F^\omega(X)$. It follows easily that every Hausdorff space embeds in a fixed point of $F$ (for instance, given any $X$, if $Y=X\coprod\mathbb{N}$ then $Y$ is naturally homeomorphic to a clopen subspace of $F(Y)$ and then $X$ embeds in $F^\omega(Y)$).

(The restriction to inclusions of clopen subspaces is probably stronger than is needed here; the key point is that you restrict to a nice enough class of inclusions such that finite products commute with sequential colimits of such inclusions.)

  • $\begingroup$ Up-voted, but this is an overkill though :-) (no, not too bad!). $\endgroup$ – Włodzimierz Holsztyński May 27 '15 at 18:34

Yet another example: $X:=\{−3,−2,−1\}∪\mathbb{Q}_+$ with the standard Euclidean topology. $\mathbb{Q}_+$ denotes the positive rationals.

I post it on an explicit request in the comments, but this is in fact a combination of the answer by Ramiro de la Vega, the finite example banned by Dominic van der Zypen and the fact that a product of a finite set by the rationals is homeomorphic to the rationals.


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