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I'm unable to understand a remark in "Two application of the method of construction by ultrapowers to analysis" by Luxemburg, which uses the ultrafilter lemma to prove the Hahn-Banach theorem, and am hoping that someone here will be able to help me.

Say that we want to extend a functional $\ell$ on a subspace $W\subset V$ which is bounded by a function $p$. Let $L$ be the set of extensions of $\ell$ to any subspace of $V$ which are bounded by $p$. For every $v\in V$, let $L_v$ be the subset of $L$ consisting of functions with domain containing $v$. This paper then shows that any ultrafilter on $L$ which contains each $L_v$ determines an extension $\tilde\ell$ of $\ell$.

At the end of the proof, it remarks that the proof demonstrates that the ultrafilter must be fixed. Presumably it means that $\{\tilde\ell\}$ is in the ultrafilter. I don't see why this should be the case and would appreciate an explanation.

I do see why the ultrafilter containing $\{\tilde\ell\}$ would yield the same extension $\tilde\ell$, but I don't see how this proves the claim.

Thanks in advance!

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    $\begingroup$ 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. $\endgroup$
    – Z. M
    Commented Nov 27, 2023 at 0:00
  • $\begingroup$ 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. $\endgroup$
    – oggius
    Commented Nov 28, 2023 at 6:12
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    $\begingroup$ 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. $\endgroup$
    – KP Hart
    Commented Nov 30, 2023 at 17:47

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I'm using the notation of the original paper.

  1. In hindsight one could have used the principal ultrafilter generated by $\{n\}$, where $n$ is the index of the extension $F$ found in the proof. But that does not imply that $\mathfrak{U}$ is that ultrafilter.

  2. As mentioned in my comment above the intersection $\bigcap\{D_x:x\in E\}$ contains all extensions with domain $E$, again in hindsight a nonempty set. That set of extensions can be infinite.

  3. Here's an example: let $E$ be $\mathbb{R}^2$ with the sum-norm $\|\cdot\|$ and $G$ just the $x$-axis. Let $f:G\to\mathbb{R}$ map $(x,0)$ to $x$. For every $a\in [-1,1]$ the functional $f_a:\mathbb{R}^2\to \mathbb{R}$ given by $f_a(x,y)=x+ay$ extends $f$ and satisfies $|f_a(x,y)|\le\|(x,y)\|$. The set of extensions is infinite and the ultrafilter in the proof could be one that extends the co-finite filter and not be guaranteed to be principal.

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  • $\begingroup$ Thanks! This is a nice proof that my suspicion is correct, and this comment in the paper was erroneous. $\endgroup$
    – oggius
    Commented Dec 2, 2023 at 19:50

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