every bijection is an homeomorphism

Question

Let M be a metric space with metric d. Assume that every bijection g: M $\rightarrow$ M is a homeomorphism. Prove that every subset of M is clopen.

Answer 1

If $M$ contains an isolated point (i.e., $a\in M$ where $\{a\}$ is open), we are done: Per suitable bijections, this makes all singleton sets open, hence all subsets are open and (as their complements are open, too) closed.

So assume $M$ has no isolated points. Pick distinct points $a,b\in M$ and let $r=\frac12 d(a,b)$. As $a$ is not isolated, we can pick $a_1\in B_r(a)\setminus\{a\}$ and then recursively $a_{n+1}\in B_{d(a,a_n)/2}(a)\setminus\{a\}$. This gives us a sequence $a_n\to a$ within the open set $B_r(a)$. Likewise, construct a sequence $b_n\to b$ within $B_r(b)$. The bijection that swaps $a_n\leftrightarrow b_n$ turns $U$ into an open set $U'$ that contains $a$, but none of the $a_n$. As $U'$ is an open neighbourhood of $a$, there exists $\rho>0$ wuch that $B_\rho(a)\subseteq U'$ - but there must be some $a_n$ in $B_\rho(a)$!