# Reconstructibility of topological spaces

Let $(X,\tau), (Y,\sigma)$ be topological spaces with $|X|$ infinite and suppose $\varphi:X\to Y$ is a bijection such that for all $x\in X$ we have that $(X\setminus\{x\}) \cong (Y\setminus\{\varphi(x)\})$.

Does this imply that $(X,\tau) \cong (Y,\sigma)$?

• Dear Dominic, this is a very interesting exercise for an elementary topology course. If $X$ has at least 3 points, the answer is "yes". Since I don't want to spoil the fun, here a hint: To check if some $U\subset X$ is open, necessarily all $U\setminus\{x\}$ must be open in $X\setminus\{x\}$. For the reverse, distinguish two cases: $U$ has at least two elements, or $X\setminus U$ has at least two elements. – Sebastian Goette Mar 20 '16 at 14:01
• @SebastianGoette: Note that an answer has been posted claiming the answer is no. So if you are sure this is right, I think it would be helpful if you would fill in the details. – Nate Eldredge Mar 20 '16 at 19:43
• All in all, there are now two very interesting exercises in one :D – მამუკა ჯიბლაძე Mar 21 '16 at 16:46
• The idea of "topological reconstruction" was explored in two papers by Max Pitz, one joint with Rolf Suabedissen: arxiv.org/abs/1311.7625, arxiv.org/abs/1501.04913 – Will Brian Mar 21 '16 at 18:53
• @მამუკა ჯიბლაძე The finite case is not obvious - for finite $T_0$ spaces it is equivalent to the unsolved reconstruction conjecture for finite ordered sets. See mathoverflow.net/a/269155/2578 . – Michał Kukieła May 6 '17 at 19:13

## 2 Answers

The answer is No''. Let $X$ be the Cantor set and let $Y = X \setminus \{p\}$, where $p$ is any element of $X$. Let $\varphi \colon X \to Y$ be any bijection. Since the Cantor set with one point removed is homeomorphic to the Cantor set with two points removed, the given condition is satisfied. However, $X$ is compact whereas $Y$ is not.

Edit: In response to the question by Loïc Teyssier, there are many ways to see that the assertion is true. One simple way is to note that if two points are removed from the Cantor set, the one-point compactification of the resulting space is compact, metrizable, zero-dimensional, and has no isolated points, and is, therefore, homeomorphic to the Cantor set. Hence, when the point at infinity is removed, the resulting space is homeomorphic to the Cantor set with a single point removed, but it is also the Cantor set with two points removed.

• "Since the Cantor set with one point removed is homeomorphic to the Cantor set with two points removed": is that true? – Loïc Teyssier Mar 20 '16 at 19:20
• The problem with the argument by Sebastian Goette is that it seems to be assuming that the homeomorphism of the spaces X \ {x} and Y \ {\phi(x)} is induced by \phi. However, that assumption is not in the original question. – Anonymous Mar 20 '16 at 21:15
• In response to the question by @LoïcTeyssier, there are many ways to see that the assertion is true. One simple way is to note that if two points are removed from the Cantor set, the one-point compactification of the resulting space is compact, metrizable, zero-dimensional, and has no isolated points, and is, therefore, homeomorphic to the Cantor set. Hence, when the point at infinity is removed, the resulting space is homeomorphic to the Cantor set with a single point removed, but it is also the Cantor set with two points removed. – Anonymous Mar 20 '16 at 21:22
• @Anonymous Indeed, I might have understood the question in the wrong way. Sorry. – Sebastian Goette Mar 20 '16 at 22:44
• @Anonymous: thanks for the precision. I assumed at some point you used the universality of the Cantor set, but couldn't quite see quickly how to make it work. – Loïc Teyssier Mar 21 '16 at 9:18

Edit. As pointed out by anonymous, the following argument assumes that the homeomorphisms $X\setminus\{x\}\cong Y\setminus\{\varphi(x)\}$ are all induced by $\varphi$. I will leave it for a while and maybe delete it later.

Recall that a subset $A\subset X\setminus\{x\}$ is open in the subspace topology if and only if there exists an open $V\subset X$ such that $A=V\cap(X\setminus\{x\})$. The subsets $V\subset X$ with $A=V\cap(X\setminus\{x\})$ are $A$ and $A\cup\{x\}$.

Consider $U\subset X$. If $U$ is open in $X$, then $U\setminus\{x\}$ is open in $X\setminus\{x\}$ for all $x\in X$.

On the other hand, let $U\setminus\{x\}$ be open in $X\setminus\{x\}$ for all $x\in X$. Assume that $U$ is not open in $X$, then $U\setminus\{x\}$ must be open in $X$ for all $x\in U$, and $U\cup\{y\}$ must be open in $X$ for all $y\notin U$.

Assume $X$ has at least three distinct elements. Then there are two cases.

If $U$ has at least two elements $x_1\ne x_2$ then $U\setminus\{x_i\}$ must be open in $X$, so $U=(U\setminus\{x_1\})\cup(U\setminus\{x_2\})$ is open, too.

If $X\setminus U$ has at least two elements $y_1\ne y_2$ then $U\cup\{y_1\}$ and $U\cup\{y_2\}$ must be open in $X$, so $U=(U\cup\{y_1\})\cap(U\cup\{y_2\})$ is open, too.