# Prime Ideal Theorem implies Hahn Banach Theorem

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!

• 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? Feb 5, 2023 at 2:54
• 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 $g\in M_P$? Feb 5, 2023 at 3:49
• @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. Feb 5, 2023 at 3:56