Timeline for When can we "displace" an ultrafilter limit with another limit?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2013 at 20:11 | vote | accept | Albert harold | ||
| Jun 27, 2013 at 23:41 | answer | added | Yemon Choi | timeline score: 2 | |
| Jun 27, 2013 at 21:03 | comment | added | Albert harold | If we can prove above notion, we see that for every $\phi\in \Delta_{\cal A}$, $\phi$-amenability is equivalent to ultra $\phi$-amenability. We say $\cal A$ is ultra $\phi$-amenable if for every ultrafilter $\cal U$, $(\cal A)_{\cal U}$ is $(\phi)_{\cal U}$-amenable. | |
| Jun 27, 2013 at 20:33 | comment | added | Albert harold | And I mean that $a_i\in \cal A$ for all $i\in F$ and $\cal U$ is a free ultrafilter on the index set $F$, and both nets are bounded. Thanks.... | |
| Jun 27, 2013 at 20:25 | comment | added | Albert harold | Yes, I want to study ultra $\phi$-amenability and ultra character amenability. I want to show that if $\phi\in \Delta_{\cal A}$, and $\cal A$ is $\phi$-amenable, then $(\cal A)_{\cal U}$ is $(\phi)_{\cal U}$-amenable, for every ultrafilter $\cal U$. But I confront to interchanging limits. | |
| Jun 27, 2013 at 18:52 | comment | added | Yemon Choi | I've just realized something - your notation is ambiguous/unclear. In order to multiply $a_i$ by $w_\alpha$ you seem to be saying that for each $i$, $a_i$ belongs to $(A_{\mathcal U})$ and not to $A$ itself. Is this correct? or do you mean that $a_i\in A$ for all $i\in F$ and $\mathcal U$ is a free ultrafilter on the index set $F$? Moreover, are both your nets bounded? | |
| Jun 27, 2013 at 18:27 | comment | added | Yemon Choi | By the way, if you used your real name, I would be more willing to help with what is clearly an attempt to write a paper/thesis on ultra-character amenability. Hiding behind a pseudonym does you no favours in this particular case; I don't see why we should do the hard work and risk not being credited in the final product. | |
| Jun 27, 2013 at 18:22 | comment | added | Yemon Choi | But you do not have uniform convergence! That is nothing to do with ultrafilters. Even if you replaced the ultrafilter limit by a limit along a sequence, you are trying to interchange two limits and BASIC analysis informs us that this might not always be possible. Your bai for ker(phi) is not necessarily a "uniform" bai. | |
| Jun 27, 2013 at 12:52 | comment | added | Albert harold | In uniformly convergence, we can displace limit operators. My goal is only inform that is this true when one of the limits is the ultrafilter limit? (because behavior of ultrafilter limit, is different from the other limits.), thank you so much. | |
| Jun 27, 2013 at 12:32 | comment | added | Albert harold | It isnt for master theses! | |
| Jun 27, 2013 at 7:49 | comment | added | Yemon Choi | If you can't prove it's true in general, why don't you try to find an example where it is false? | |
| Jun 27, 2013 at 7:43 | comment | added | Yemon Choi | Before asking a question on MO, you should demonstrate that you have made a serious attempt to answer the question on your own. We are not in the business of writing people's master's theses for them | |
| Jun 27, 2013 at 7:28 | history | edited | Albert harold | CC BY-SA 3.0 |
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| Jun 27, 2013 at 7:22 | comment | added | Albert harold | In which cases this notion can be true? | |
| Jun 27, 2013 at 6:55 | review | Close votes | |||
| Jul 10, 2013 at 3:01 | |||||
| Jun 27, 2013 at 6:38 | comment | added | Yemon Choi | What reason do you have to believe that one can interchange limits? In general, in analysis, one cannot do this, so you should give some evidence or special cases which show why this might be true. | |
| Jun 27, 2013 at 6:36 | comment | added | Yemon Choi | Doing ultra-character amenability, are we? | |
| Jun 27, 2013 at 6:06 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
added 21 characters in body; edited title
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| Jun 27, 2013 at 6:01 | history | asked | Albert harold | CC BY-SA 3.0 |