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Formally question:

Consider an increasing function $f:\mathbb{R}\to\mathbb{R}$ (not necessarily continuous), if $I\subset\mathbb{R}$ is connected, then $f^{-1}(I)$ is connected. How I can to proof that question?

Andhreyux Muñoz Cid
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos May 19 '22 at 01:57
  • My context is that the connectedness properties under continuous functions are appropriate and under certain discussions I was told that this was true. But under the hypothesis that it is an increasing function it doesn't make sense to me. I have been thinking of supposing that the pre image is not connected, but the hypothesis of being increasing does not help me or I do not know how to use it. – Andhreyux Muñoz Cid May 19 '22 at 02:05
  • Connected subsets of the real line must be intervals. – Randall May 19 '22 at 02:06
  • That is true @Randall, but how to use f increasing in this case? – Andhreyux Muñoz Cid May 19 '22 at 02:09
  • Preimages of convex sets are convex. – Ulli May 19 '22 at 05:44

1 Answers1

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If $f$ is (strictly) increasing, then $f^{-1}$ is continuous, and since continuous functions map connected to connected, the result follows.

Jamie Alizadeh
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    That is a great hint , I’m going to write a formal proof. Thank you – Andhreyux Muñoz Cid May 19 '22 at 02:52
  • $f^{-1}$ only exists, if $f$ is bijective. However the above also holds, if $f$ is not injective or surjective, see my above comment – Ulli May 19 '22 at 05:47
  • Strictly increasing real functions are invertible – Jamie Alizadeh May 19 '22 at 13:40
  • Certainly not, if they are not surjective. Moreover, the OP did not ask for strictly increasing functions. – Ulli May 19 '22 at 13:51
  • Strictly increasing implies surjective. In my answer I assumed that $f$ is strictly increasing and OP verified it - so we can assume OP meant strictly increasing. – Jamie Alizadeh May 19 '22 at 15:15
  • Oh, that's really strange! Just to understand your point: for instance, $\textrm{exp}: \bf{R} \rightarrow \bf{R}$ is not strictly increasing in your opinion? Well, if it's fine for the OP, then it's ok. Just wondering, if someone else stumbles over this question ... – Ulli May 19 '22 at 15:57
  • Ah, strictly increasing doesn't necessarily imply surjective you're right. I should have said it implies injectivity and injectivity implies invertibility from the domain to the image. – Jamie Alizadeh May 19 '22 at 16:38