Timeline for Continuity/measurability of a complicated extension of a family of continuous functions
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2011 at 16:32 | comment | added | Yulia Kuznetsova | Maybe one can modify it to be non-measurable. The task is to find such a chain $F_i$ that: every $F_i$ is closed; $E=\cup F_i$ is nowhere dense and not Borel. Then the counterexample will be given by $f_i|_{F_i}=0$, $f_i|_{X\setminus E}=1$ where $\\{0,1\\}$ is the Sierpinski space with 1 open. | |
| Dec 7, 2011 at 20:05 | comment | added | Salvo Tringali | @Yulia. Yes, I need it to be linearly ordered. | |
| Dec 7, 2011 at 12:48 | comment | added | Yulia Kuznetsova | @Salvo - if you do not require $X_i$ to be a chain (i.e. linearly ordered), it can be easily modified to yield a non-measurable function. Are they really a chain in your applications? | |
| Dec 6, 2011 at 15:20 | comment | added | Salvo Tringali | @Yulia. Too bad that your nice counterexample doesn't work also for the 2nd question. In any case, I'm going to edit the OP and add that you replied Q1 in the negative. | |
| Dec 5, 2011 at 18:12 | comment | added | Yulia Kuznetsova | Emil - corrected, thank you. I didn't want 0 to be dense, I wanted to explain briefly that the only open sets are $\emptyset,\{1\}$ and $Y$. | |
| Dec 5, 2011 at 18:11 | history | edited | Yulia Kuznetsova | CC BY-SA 3.0 |
deleted 32 characters in body
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| Dec 5, 2011 at 17:39 | comment | added | Emil Jeřábek | Also, the $U_k$ are not used anywhere, and there is a missing $-$ in the definition of $X_i$. | |
| Dec 5, 2011 at 17:36 | comment | added | Emil Jeřábek | If $\{1\}$ is open, then $\{0\}$ is closed, hence it can’t be dense. But you don’t really need the latter in the argument, it works with $Y$ being the Sierpiński space. | |
| Dec 5, 2011 at 17:04 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
fix markup
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| Dec 5, 2011 at 16:58 | comment | added | Yulia Kuznetsova | Yes, I know how to type TeX... Done. This time it worked, but before the same text didn't want to show after multiple edits. | |
| Dec 5, 2011 at 16:57 | history | edited | Yulia Kuznetsova | CC BY-SA 3.0 |
added 170 characters in body
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| Dec 5, 2011 at 16:33 | comment | added | Yulia Kuznetsova | Sorry, need to correct this: obviously $X_i$ are not dense like this, one should put $X_i = [0,1] \setminus \{2^k: k>i\}$. | |
| Dec 5, 2011 at 16:27 | comment | added | Yulia Kuznetsova | (continue because of problems with TeX) .. and let $X_i=\{0\}\cup [2^{-k},1]$. Put $f_i(0)=1$, $f_i(2^{-k})=0$ for all $k$, and $f_i(x)=1$ for other $x$. Then every $f_i$ is continuous but $f$ is not. Right? | |
| Dec 5, 2011 at 16:20 | history | answered | Yulia Kuznetsova | CC BY-SA 3.0 |