# Shift of first category set in a compact metric space.

Question. If $X$ is a homogeneous compact metric space, and $F=\bigcup _{n\in\omega}F_n$ is a countable union of closed nowhere dense subsets of $X$, then is there a homeomorphism $\varphi:X\to X$ such that $F\cap\varphi[F]=\varnothing$?

This is my own question that I have not been able to answer.

If the answer is yes then there might be some interesting applications, but I think it is interesting in its own right.

EDIT. I was originally thinking of the Cantor set. Then I over-generalized. Clearly the $X$ must be homogeneous.

• Take X to be the closed unit interval. Now think of its endpoints. – Moishe Kohan Jun 9 '16 at 0:16
• @studiosus yes homogeneous is needed. see edit – Forever Mozart Jun 9 '16 at 0:22
• Oh, I see. Then take projective plane and a projective line inside. – Moishe Kohan Jun 9 '16 at 0:35
• @studiosus can you explain more? (in an answer would be best) – Forever Mozart Jun 9 '16 at 0:56
• Why don't you accept Eric's answer? This will move the question to "answered" lit. – Moishe Kohan Jun 12 '16 at 13:42

To elaborate on studiosus's comment, the projective plane is a counterexample. Let $X=\mathbb{R}P^2$ ($\mathbb{C}P^2$ would also work) and let $F\subset X$ be the line at infinity. Then $F$ is a closed subset of $X$ with empty interior, and in particular is first category. The complement $X\setminus F$ is homeomorphic to $\mathbb{R}^2$, and $F\cong \mathbb{R}P^1$ is homeomorphic to a circle $S^1$. Moreover, the inclusion map $F\to X$ is not nullhomotopic (for instance, it induces a surjection $\mathbb{Z}\cong \pi_1(F)\to\pi_1(X)\cong\mathbb{Z}/2$).
Now suppose there were a homeomorphism $\varphi:X\to X$ such that $\varphi(F)\cap F=\emptyset$. Then $\varphi(F)\subseteq X\setminus F\cong \mathbb{R}^2$. Since $\mathbb{R}^2$ is contractible, this would mean the inclusion map $\varphi(F)\to X$ is nullhomotopic. But this would imply the inclusion $F\to X$ is also nullhomotopic, since $\varphi$ is a homeomorphism. Thus no such $\varphi$ can exist.
• For a similar example you may find easier to think about, you could let $X=S^1\times S^1$ (thought of as a quotient of a square) and $F=S^1\vee S^1$ (the image of the boundary of the square under the quotient map). The same argument applies: the inclusion map $i:F\to X$ is not nullhomotopic, but $X\setminus F$ is just the interior of a square and hence is contractible. – Eric Wofsey Jun 9 '16 at 1:21