3
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ I took the liberty of eliminating the notation collision at $x$. $\endgroup$ – Benoît Kloeckner May 15 '17 at 11:50
7
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.