Timeline for Do multiplicative Banach limits exist?
Current License: CC BY-SA 4.0
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| Apr 7, 2020 at 8:29 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 8:24 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 8:24 | vote | accept | aduh | ||
| Apr 7, 2020 at 8:08 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 8:03 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 7:56 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 7:51 | comment | added | Martin Sleziak | Re: Why $\mathcal U$-limit exists: Since $\mathcal U$ is an ultrafilter and the values are all on some bounded interval, which is a compact space. A proof that in compact space limit along an ultrafilter exists is given in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. (And probably in many other resources.) | |
| Apr 7, 2020 at 7:43 | comment | added | aduh | Also, I wonder if, in general, the existence of multiplicative Banach limits requires the ultrafilter lemma (which is stronger than Hahn-Banach). | |
| Apr 7, 2020 at 7:41 | comment | added | aduh | I'm not sure I see why the $\mathcal U$-limit exists. Can you explain? | |
| Apr 7, 2020 at 7:41 | comment | added | aduh | Thanks. My definition comes from Howard & Rubin's Consequences of the Axiom of Choice and Schechter's Handbook of Analysis. It's just a bit more general than yours, so I don't think there are any issues there. | |
| Apr 7, 2020 at 7:36 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 7:23 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 7, 2020 at 7:06 | history | answered | Martin Sleziak | CC BY-SA 4.0 |