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Let $X$ be a complete real inner-product space and $f:X \to X$ be a bijection which maps connected sets to connected sets ; then is it necessarily true that $f^{-1}$ also maps connected sets to connected sets ?

Vim
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    Here (http://math.stackexchange.com/questions/952466/does-there-exist-a-bijection-of-mathbbrn-with-itself-such-that-the-forward), the same question has been asked only for $\Bbb{R}^n$ and did not receive a satisfactory answer. So if at all, there might be an infinite dimensional counterexample (or someone finds an answer to the other question too :) ) – PhoemueX Aug 02 '15 at 07:06
  • @Saun Dev: Copied from my comment on question 952466.... Consider $f:\mathbb{R}^2\to\mathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+\sin(\pi/s)$ for $s\neq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected. – Alan U. Kennington Aug 29 '15 at 21:05
  • In my previous comment, I meant $f(s,t)=(s,t+\sin(\pi/s))$ for $s\neq0$. – Alan U. Kennington Aug 30 '15 at 00:22

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