Timeline for Every function on reals a sum of two surjective real functions?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| S Apr 26 at 1:22 | history | suggested | Cameron Buie | CC BY-SA 4.0 |
clarified terminology
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| Apr 25 at 23:29 | review | Suggested edits | |||
| S Apr 26 at 1:22 | |||||
| Jan 5 at 4:56 | vote | accept | vidyarthi | ||
| Jan 4 at 19:22 | comment | added | Joel David Hamkins | Yes, or just look at the parity of the first symbol after it becomes well formed, and then omit that symbol (which is a little closer to your original E O idea). | |
| Jan 4 at 18:51 | comment | added | Nick S | @JoelDavidHamkins In the Conway 13 function, pick $g(x)=r$ or $h(x)=r$ only if it is a well-formed integer after an odd/even number of places. Or if the number of "special" characters is odd/even. Otherwise chose $g+h=f$. | |
| Jan 4 at 18:50 | comment | added | Joel David Hamkins | Actually the E O idea didn't quite achieve that, but one can achieve it. | |
| Jan 4 at 18:46 | comment | added | Joel David Hamkins | If you use sets that are disjoint and both size continuum on every interval (which your previous E O in effect did), you can use something like the Conway 13 function idea to make both $g$ and $h$ onto from every interval, which is a bit more than just IVT. | |
| Jan 4 at 18:45 | comment | added | Nick S | @JoelDavidHamkins Done :) | |
| Jan 4 at 18:45 | comment | added | Nick S | @vidyarthi Yes, you can take any two disjoint subsets of $\mathbb R$ such that there is an onto function from each of them to $\mathbb R$. The construction only works if the two sets are disjoint, that's why I picked those. See the new solution....... | |
| Jan 4 at 18:43 | history | edited | Nick S | CC BY-SA 4.0 |
deleted 673 characters in body
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| Jan 4 at 17:33 | comment | added | Joel David Hamkins | I suggest you edit to omit the O, E stuff and just keep the simplification argument, which seems very clear. | |
| Jan 4 at 17:25 | comment | added | vidyarthi | Ok, but I did not get what the use of the sets $O$ and $E$ are. Like, similar to your construction, I could take any two surjective functions on some restricted subsets of $\mathbb{R}$ and then perform similar construction/extension right | |
| Jan 4 at 17:21 | history | edited | Nick S | CC BY-SA 4.0 |
added 290 characters in body
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| Jan 4 at 17:20 | comment | added | Nick S | @vidyarthi No. Note that on $E$, the function $h$ is defined as $h=f-g$, which allows $f$ be anything.... Just to make this point crystal clear, when $f=0$ this construction produces $h=-g$... This works, even if $f$ is not surjective. | |
| Jan 4 at 17:17 | comment | added | vidyarthi | But then, every such $f$ will be surjective right? | |
| Jan 4 at 17:11 | history | answered | Nick S | CC BY-SA 4.0 |