Timeline for Hahn-Banach theorem and ultrafilter lemma
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2023 at 19:49 | vote | accept | oggius | ||
| Nov 30, 2023 at 21:04 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 30, 2023 at 20:23 | answer | added | KP Hart | timeline score: 2 | |
| Nov 30, 2023 at 17:47 | comment | added | KP Hart | What the proof shows, of course, is that the intersection of the family $\{D_x:x\in E\}$ is nonempty, as it contains the extension $F$. It seems unlikely that $A=\bigcap\{D_x:x\in E\}$ contains only finitely many elements wiith domain $E$. If $A$ is infinite you can add the cofinite filter on $A$ to $\{D_x:x\in E\}$ and thus get an(other) ultrafilter $\mathcal{V}$ that also provides an extension but that is not principal. | |
| Nov 28, 2023 at 6:12 | comment | added | oggius | Thanks. I'm not having trouble following Luxemburg's proof of Hahn-Banach, just his remark afterwards which I think might be erroneous, but your proof was neat too. | |
| Nov 27, 2023 at 0:00 | comment | added | Z. M | I am not sure about that paper, but it is not difficult to prove Hahn–Banach by Tychonoff's theorem for compact Hausdorff spaces, which is equivalent to the ultrafilter lemma. Indeed, let $E$ be the set of functions $f\colon V\to\mathbb R$ such that $\lvert f(v)\rvert\le p(v)$ for every $v\in V$. Now for every subspace $U$ of $W$ containing $W$ such that $U/W$ is finite dimensional, let $E_U\subset E$ denote the subset of $f$ such that $f\rvert_U$ is linear and $f\rvert_W=\ell$, a non-empty closed in $E$, and any $f$ in the intersection among all $U$, which is non-empty by compactness, is OK. | |
| Nov 26, 2023 at 19:21 | history | asked | oggius | CC BY-SA 4.0 |