Timeline for Bijective function on a dense set
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2011 at 3:49 | vote | accept | FelipeG | ||
| Nov 1, 2011 at 3:49 | |||||
| Nov 1, 2011 at 3:49 | vote | accept | FelipeG | ||
| Nov 1, 2011 at 3:49 | |||||
| Oct 31, 2011 at 20:47 | history | edited | KP Hart | CC BY-SA 3.0 |
added one-diensional example.
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| Oct 31, 2011 at 18:51 | history | edited | KP Hart | CC BY-SA 3.0 |
added 2 characters in body; added 132 characters in body
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| Oct 31, 2011 at 18:40 | history | edited | KP Hart | CC BY-SA 3.0 |
Added a description of a construction of $f$
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| Oct 31, 2011 at 17:59 | comment | added | fedja | Ah, right you are! I somehow missed that part of the condition entirely. Shame on me! | |
| Oct 31, 2011 at 14:30 | comment | added | Joel David Hamkins | Indeed, for the same reason, no point of $D$ can share in the non-injectivity of $f$. So there is no way to make your function work on $[-1,1]$, as every non-zero point is involved in a failure of injectivity. | |
| Oct 31, 2011 at 13:26 | comment | added | Joel David Hamkins | Fedja, if you only take some of the pre-image, then you won't have $f^{-1}(D)=D$. | |
| Oct 31, 2011 at 13:20 | comment | added | fedja | ---I think I can prove that if X=[0,1] (or 1-dimensional) then the claim of FelipeG is true--- That's unlikely. Take $f:x\mapsto 2x^2-1$ on $[-1,1]$ and take some dense set of transcendental algebraicly independent numbers. Now, for each number take its full forward orbit and some its backward orbit (say, choosing the preimage from $[0,1]$ at each step). Clearly, all numbers you get will be different, so $f$ is one to one on the resulting dense set $D$. A similar construction should work in all but very degenerate metric spaces. | |
| Oct 31, 2011 at 8:03 | comment | added | ccarminat | I think your counterexample contains a bug (probably the problem is in the continuous extension). In fact I think I can prove that if X=[0,1] (or 1-dimensional) then the claim of FelipeG is true. | |
| Oct 31, 2011 at 0:00 | comment | added | Guillaume Brunerie | I don't understand how you can extend your function to $[0,1]$, you will need uniform continuity. | |
| Oct 30, 2011 at 22:32 | history | answered | KP Hart | CC BY-SA 3.0 |