mathlearner98 Asks:
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
\leq p
$ 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 $$N_P=\{z\in Z
\in \text{dom}(z)\}$$ $N_P$'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!
SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.
