If $f:X \to X$ is a continuous bijection and every point has finite orbit, is $f^{-1}$ continuous?

Solution 1:

This seems to be a counterexample. Take $X = \mathbb{N}^2 \cup \{\infty\}$ with the metric where $\rho((m,n), \infty) = 1/n$ and $\rho((m,n), (p, q)) = 1/n + 1/q$ when $(m,n) \ne (p,q)$. Define $f(\infty) = \infty$ and $$ f((m,n)) = \cases{ (m, n-1) & if $2 \le n \le m$, \\ (m, m) & if $n=1$, \\ (m, n) & otherwise. } $$ For every $m \in \mathbb{N}$ the set $\{ (m, 1), \dots, (m, m) \}$ is an finite orbit and any point not in such a set is fixed. Since $(\mathbb{N}^2, \rho)$ is discrete, we only need to consider continuity at $\infty$. It is easy to verify that $\rho(f(m,n), \infty) \le 2\rho((m,n), \infty)$, so $f$ is continuous. On the other hand, every neighbourhood of $\infty$ contains a point $(m,m)$ for some value of $m$ and $\rho(f^{-1}(m,m), \infty) = 1$, hence $f^{-1}$ is not continuous.

For a slightly more natural counterexample, you could construct a vector space automorphism of $\mathbb{R}^\infty$ in a similar way, for example such that $$\begin{eqnarray} Te_1 &=& 2e_2, Te_2 = \frac{e_1}{2}, \\ Te_3 &=& 2e_4, Te_4 = 2e_5, Te_5 = \frac{e_3}{4} \\ \text{etc.}&& \end{eqnarray} $$ Clearly $T$ is a bounded linear operator w.r.t. the Euclidean norm but its inverse is not bounded. To see that it has finite orbits, note that the orbits of basis vectors are finite, say $T^{k_i} e_i = e_i$, and it follows for $x = \sum_{i=1}^n c_ie_i$ that $T^K x = x$, where $K = \prod_{i=1}^n k_i$.