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I came upon this when trying to solve a similar problem first: Open maps which are not continuous(1), which is essentially my problem without requiring the map to be bijective.

To my knowledge, there are a bunch constructions satisfying the weaker constraints: Conway base 13 function(https://en.wikipedia.org/wiki/Conway_base_13_function), a cool one using Riemann Series Theorem (see (1)), and basically all strongly Darboux functions.

The problem is that all these constructions are not bijective, and I'm looking for a bijective example. Immediately this disqualifies all strongly Darboux functions, as they are not bijective on any open set, and this is my progress so far.

Stone Echo
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1 Answers1

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The answer to the question in the title is negative.

We know that if a function $f:I \to \Bbb{R}$ is continuous and injective then its inverse $f^{-1}:f(I) \to I$ is also continuous.

Here $I$ denotes an interval .

Now since $f$ is open and a bijection we have that $f^{-1}$ is a continuous function,so $(f^{-1})^{-1}$ is also continuous.

Marios Gretsas
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  • I suggest that you add that $I$ is an interval. Otherwise, the statement that you are using is false. – José Carlos Santos Sep 19 '21 at 07:22
  • @JoséCarlosSantos..thank you for the correction..i edited.. i did not mention anything because with $I$ we usualy denote intervals – Marios Gretsas Sep 19 '21 at 07:24
  • Would you further explain why open and onto suggests continuous inverse? I'm only starting to learn multivariable calculus (which is when my textbook introduces open/close sets), and I might not be aware of some facts. – Stone Echo Sep 19 '21 at 07:27
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    @StoneEcho a function $g$ is continuous iff the inverse image of an open set under $g$ is open – Marios Gretsas Sep 19 '21 at 07:28
  • @MariosGretsas Sure enough. I guess I was caught by the thought of finding a counter-example and overlooked such basic conclusion. Thanks a lot! – Stone Echo Sep 19 '21 at 07:39
  • I am topology newbie. What is wrong with the following attempt at a counter example? If $(1/2)$ divides $x$, then $f(x) = x.$ Otherwise, suppose that $x = \lfloor x \rfloor + r_x ~: ~0 < r_x < (1/2) ~$ or $~(1/2) < r_x < 1.$ Then, define $f(x) = \lfloor x\rfloor + (1 - r_x).$ $f$ is bijection to all of $\Bbb{R}$, $f$ not continuous at any integer, and $f$ is its own inverse. – user2661923 Sep 19 '21 at 08:32
  • @user2661923 Is $f$ an open map? – Marios Gretsas Sep 19 '21 at 08:34
  • @MariosGretsas Unsure, but I think so, since $\Bbb{R}$ is an open set, and $f$ is surjective from $\Bbb{R}$ to $\Bbb{R}$. – user2661923 Sep 19 '21 at 08:36
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    @user2661923 an ''open map'' means the $f$ sends every open set to an open set..not only $\Bbb{R}$ – Marios Gretsas Sep 19 '21 at 08:37
  • @MariosGretsas Oh, okay. Thanks for clarification. – user2661923 Sep 19 '21 at 08:38
  • A continuous $f : I \to \mathbb R$ need not be injective, thus $f^{-1}$ in general does not exist. If $f$ is injective, then you have $f^{-1} : f(I) \to I$. And don't you invoke a result ("we know ...") that the OP wants to have explained? – Paul Frost Sep 19 '21 at 10:41
  • @PaulFrost the O.P asks for bijective open maps...also i pressumed that since the O.P has learned open and closed sets,then has learn also about continuity in terms of open sets.. – Marios Gretsas Sep 19 '21 at 10:43
  • Then you should edit your question to include that $f$ is a bijection. Anyway, the range of $f^{-1}$ is $I$. – Paul Frost Sep 19 '21 at 10:44
  • @PaulFrost i edited,thank you for the correction – Marios Gretsas Sep 19 '21 at 10:46