Timeline for Bijection $f: \mathbb R^n \to \mathbb R^n$ that maps connected onto connected sets must map closed connected onto closed connected sets?
Current License: CC BY-SA 4.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| S Sep 8, 2018 at 11:41 | history | bounty ended | Right | ||
| S Sep 8, 2018 at 11:41 | history | notice removed | Right | ||
| Sep 6, 2018 at 5:36 | answer | added | Taras Banakh | timeline score: 14 | |
| Sep 4, 2018 at 20:23 | history | edited | Right | CC BY-SA 4.0 |
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| Sep 4, 2018 at 20:14 | history | edited | Right | CC BY-SA 4.0 |
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| Sep 3, 2018 at 4:15 | comment | added | Taras Banakh | @Right I had in mind that if you resolve your question affirmatively, then you can apply the theorem of Jones and will resolve the question of Wong. Please look at the paper of Jones and the formulation of its main result. | |
| Sep 2, 2018 at 23:17 | history | edited | Right | CC BY-SA 4.0 |
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| S Sep 2, 2018 at 23:12 | history | bounty started | Right | ||
| S Sep 2, 2018 at 23:12 | history | notice added | Right | Canonical answer required | |
| Sep 2, 2018 at 23:11 | history | edited | Right | CC BY-SA 4.0 |
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| Sep 2, 2018 at 23:08 | comment | added | Right | @TarasBanakh Do you mean that I can safely assume that f maps closed connected subsets onto closed connected subsets to resolve a question of Wong? | |
| Sep 2, 2018 at 21:04 | comment | added | Taras Banakh | By a result of Jones (ams.org/journals/proc/1968-019-01/S0002-9939-1968-0222857-1/…) the question of @Right is quivalent to the question of Wong. | |
| Aug 27, 2018 at 5:51 | comment | added | Brevan Ellefsen | @NateEldredge this is what I meant when I gave the closed map remark above - I had this argument in my head, and wrote too quickly without explanation. Of course, it might very well turn out that we can independently prove $f$ is a closed map or even merely that its restriction to connected sets is a closed map, and thus bypass the need for a positive answer to Willie Wong's question. I've now deleted my unclear comments above, and repeat the link for those interested math.stackexchange.com/questions/949168/… | |
| Aug 27, 2018 at 5:49 | comment | added | Brevan Ellefsen | @NateEldredge The linked post addresses this question as follows: We know $f$ is a bijection that maps connected sets to connected sets. We wish to show $f$ is also a homeomorphism, as then $f$ is a closed map and this question is trivial. Due to a property of homeomorphisms, we can show $f$ cannot be a homeomorphism unless both $f$ and $f^{-1}$ map connected sets to connected sets, and under this hypothesis we can show $f$ is indeed a homeomorphism. We are thus dependent on the answer to Willie Wong's original post, which is still open. | |
| Aug 27, 2018 at 5:46 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 27, 2018 at 5:39 | comment | added | Brevan Ellefsen | @NateEldredge right you are about the closed map remark. Skimmed through too quickly and thought too little. Regardless, the link I gave in my comment above should still be applicable. The link answers the question in the second quote box. The question in the first quote box is still an open problem | |
| Aug 27, 2018 at 5:39 | comment | added | Nate Eldredge | @BrevanEllefsen: The link addresses the question in the first quote box, which as you say has been answered by Willie. It's included for context, but the point of this post is the question in the second box. | |
| Aug 27, 2018 at 5:36 | comment | added | Nate Eldredge | @BrevanEllefsen: If $T$ is a non-connected closed set, then we are not asking for $f(T)$ to be closed. | |
| Aug 27, 2018 at 3:20 | history | asked | Right | CC BY-SA 4.0 |