Timeline for A subcontinuous function, which is not continuous
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2020 at 16:10 | comment | added | Gro-Tsen | What I called $u_k$ obviously wasn't what was called $u_k$ in the question: I changed this to $t_k$ to avoid confusion. Terms with $u_k<0$ are unproblematic, all terms will be $\leq 1$ eventually, and terms between $0$ and $1$ can be written $t/n+(1-t)/(n+1)$ with $0\leq t\leq 1$ so their image by $f is $te_n+(1-t)e_{n+1}$. | |
| Jul 27, 2020 at 16:04 | history | edited | Gro-Tsen | CC BY-SA 4.0 |
rename u_k to t_k to avoid conflict of notation with question
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| Jul 27, 2020 at 15:42 | comment | added | Motaka | The sequence $u_k$ must be general (and converges to $0$), I don't know why you assume that $0\leq u_k\leq 1$? | |
| Jul 18, 2020 at 10:24 | vote | accept | Motaka | ||
| Jul 18, 2020 at 10:02 | history | edited | Gro-Tsen | CC BY-SA 4.0 |
fix typo/thinko
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| Jul 18, 2020 at 9:35 | comment | added | username | That means if $x\in[(n+1]^{-1},n^{-1}]$ then $$f(x) = f(\frac1n) +(f(\frac1{n+1}) -f(\frac1n))\frac{x-\frac1n}{\frac1{n+1} - \frac1n} $$ | |
| Jul 18, 2020 at 9:28 | comment | added | Motaka | What the you mean exactly by : interpolating linearly between... | |
| Jul 18, 2020 at 9:24 | history | answered | Gro-Tsen | CC BY-SA 4.0 |