I am reading Jech's Axiom of Choice, and there is this exercise: chapter 2 Problem 19:

Show that the Hahn-Banach Theorem follows from the Prime Ideal Theorem.

I came up with a (possibly wrong) proof, and my idea is essentially to mimic the proof of the implication "Prime Ideal Theorem $\implies$ Consistency Principle for Binary Mess" (the proof of this implication is in Jech's book chapter 2). Because I am a novice in this field, I really am not sure whether or not I am missing something, so I would really appreciate if you can point out if I am missing something.

Let $V$ be a real vector space and $X\subset V$ be a subspace. Let $p$ be a sublinear functional on $V$ and let $f$ be a linear functional on $X$ such that $f(x)\leq p(x)$ for all $x\in X$. Using Ultra Filter Theorem (which is equivalent to the Prime Ideal Theorem) we will show that there is a linear functional $F$ on $V$ such that $F(x)\leq p(x)$ for all $x\in V$ and $F\restriction_X=f$. We will use this lemma without proof (proof can be found at page 158 Folland):

Lemma A: For every $x\in V$, there is a linear functional $g$ on $X+\mathbb{R}x$ extending $f$ with $g(y)\leq p(y)$ for all $y\in X+\mathbb{R}x$.

Let $I$ be the set of all finite dimensional subspaces of $V$. Let $$M:=\{g:g\text{ is a linear functional on some }P\in I\text{ compatible with }f\text{ and }g(x)\leq p(x)\text{ for all }x\in P\}$$ From Lemma A, for each $P\in I$, there is some $g\in M$ with $\text{dom}(g)=P$. For each $P\in I$, let $M_P$ denote the set of all $g\in M$ such that $\text{dom}(g)=P$. We take $Z$ be the set of all functions $z$ such that:

a) $\text{dom}(z)\subset I$;

b) $z(P)\in M_P$ for each $P\in \text{dom}(z)$;

c) for any $P,Q\in \text{dom}(z)$, the functions $g_1=z(P)$ and $g_2=z(Q)$ are compatible.

Let $\mathbb{F}$ be the filter over $Z$ generated by the sets $$N_P=\{z\in Z:P\in \text{dom}(z)\}$$ $N_P$'s indeed do generate a filter over $Z$ because for $P_1,...,P_n\in I$, we know $N_{P_1}\cap \cdots\cap N_{P_n}$ is non-empty, as $P_1+\cdots+P_n$ is still a finite dimensional subspace of $V$, and hence there is some linear functional $g$ on $P_1+\cdots+P_n$ such that is compatible with $f$ and $g(x)\leq p(x)$ for all $x\in P_1+\cdots+P_n$. Then $z:=\{(P_i\to g\restriction_{P_i}):1\leq i\leq n\}$ is inside $N_{P_1}\cap\cdots\cap N_{P_n}$.

By the Ultra Filter Theorem, we may let $U$ be an ultra filter over $Z$ that extends $\mathbb{F}$. Now, fix $P\in I$. For each $g\in M_P$, denote $O_g:=\{z\in Z:z(P)=g\}$. Then note that $N_P=\bigcup_{g\in M_P}O_g$ and that $O_g$'s are pair-wise disjoint. Also, each $O_g$ is non-empty. So from the fact that $U$ is an ultra filter, there exists a unique $g\in M_P$ such that $O_g\in U$. Hence, for an arbitrary $P\in I$, we found a unique choice of a linear functional (which we denote by) $g_P$ defined on $P\subset V$.

We claim that $g_P$'s for $P\in I$ are pair-wise compatible. For $P,Q\in I$, we know $O_{g_P},O_{g_Q}\in U$. Hence, $O_{g_P}\cap O_{g_Q}\neq\emptyset$. This means $g_P$ and $g_Q$ are compatible. Now, if we take $F:=\bigcup_{P\in I}g_P$, then $F$ is the desired linear functional on $V$.

Thank you in advance for any help!

  • $\begingroup$ Are you saying that since $N_P\in U$ and $N_P=\bigcup_{g\in M_P}O_g$, one of $O_g$ has to be in $U$? Why is that true since $M_P$ is infinite? $\endgroup$ Feb 5, 2023 at 2:54
  • $\begingroup$ I think your point may be a crucial problem in my proof. Do you mean it is possible that $O_g^c\in U$ for all $g\in M_P$? $\endgroup$ Feb 5, 2023 at 3:49
  • $\begingroup$ @newaccount Yes, that was what I thought, but I think you are right. I am trying to look for a way to fix this issue now. $\endgroup$ Feb 5, 2023 at 3:56


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