Timeline for Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ with injective restriction $f|\mathbb Q^\omega$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2018 at 21:26 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 15 characters in body
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| Sep 17, 2018 at 17:26 | comment | added | Gro-Tsen | Oh yes of course, silly question. Thanks. | |
| Sep 17, 2018 at 15:38 | comment | added | Fedor Petrov | @Gro-Tsen by injectivity it contains two different points and by continuity on the segment between them we get the whole segment | |
| Sep 17, 2018 at 14:59 | comment | added | Gro-Tsen | Wait, why does for any $p_1$ the set $f(p_1\times\mathbb{R}^{\omega-1})$ have non-empty interior? Isn't the first projection $f\colon(p_n)_{n\in\omega}\,\mapsto p_1$ a counterexample? Or are you using injectivity of $f$ on $\mathbb{Q}^\omega$, but how? (I remember convincing myself that your argument was correct, but now I all confused.) | |
| Sep 17, 2018 at 10:30 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
edited body
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| Sep 3, 2017 at 6:47 | vote | accept | Taras Banakh | ||
| Sep 3, 2017 at 6:47 | comment | added | Taras Banakh | Thank you for the nice and short solution. I also arrived to negative answer and started to write it (using Baire Theorem), but your solution is much shoter and nice. | |
| Sep 3, 2017 at 6:10 | history | answered | Fedor Petrov | CC BY-SA 3.0 |