Let $(X,\tau)$ be a topological space. For $x,y\in X$ we set $$\Delta(x,y) = \{(z,z):z\in X\}\cup \{(x,y),(y,x)\}.$$

Is there a connected Hausdorff space $(X,\tau)$ with more than $1$ point such that for all $x,y\in X$ we have $X\cong X/\Delta(x,y)$ with respect to the quotient topology?

  • $\begingroup$ I took the liberty of eliminating the notation collision at $x$. $\endgroup$ – Benoît Kloeckner May 15 '17 at 11:50

Start with the space $X_0=S^1$ (any connected manifold will do), and choose a basepoint in $X_0$. Construct a sequence of pointed spaces recursively as follows: $$X_{n+1}=\bigvee_{x,y\in X_n, x\ne y} X_n/ \Delta(x,y).$$ Let $X=\bigvee_{n=1}^\infty X_n$.

Then $X$ is connected and Hausdorff. Moreover, $X$ is naturally identified with a wedge sum with the following properties: 1. For each wedge summand, there are infinitely many other summands homeomorphic to it, 2. for any two points in a wedge summand $A$, there is a wedge summand homeomorphic to the quotient of $A$ by the two points, 3. For any two points belonging to different wedge summands $A, B$, there is a wedge summand homeomorphic to the quotient of $A\vee B$ identifying these points. From here it is easy to conclude that $X$ has the property that you want.

  • $\begingroup$ I wonder if such a thing could be compact. I suspect not. $\endgroup$ – Jeff Strom May 15 '17 at 11:57
  • 2
    $\begingroup$ I think maybe you can construct a compact example by modifying the above construction where instead of taking wedge sum, you embed things in ${\mathbb R}^n$ in the style of Hawaiian earring. More precisely, choose a continuous injective map $X\subset {\mathbb R}^2$ whose image is closed and bounded. I think it is not difficult to see that a map of this kind exists (may be easier to visualize with ${\mathbb R}^3$ instead of ${\mathbb R}^2$), whose image still has the required property. $\endgroup$ – Gregory Arone May 15 '17 at 12:48

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