This problem was posed on Math StackExchange some time ago, but it did not garner any solutions there. I think that it is interesting enough to be posed here on Math Overflow, so here it goes.


Let $ \mathcal{A} $ be a unital Banach algebra over $ \mathbb{C} $, with $ \mathbf{1}\_{\mathcal{A}} $ denoting the identity of $ \mathcal{A} $. For each $ a \in \mathcal{A} $, define the spectrum of $ a $ to be the following subset of $ \mathbb{C} $:


$$
{\sigma\_{\mathcal{A}}}(a) \stackrel{\text{def}}{=} \lbrace \lambda \in \mathbb{C} ~|~ \text{$ a - \lambda \cdot \mathbf{1}\_{\mathcal{A}} $ is not invertible} \rbrace.
$$


With the aid of the Hahn-Banach Theorem and Liouville's Theorem from complex analysis, one can prove the well-known result that $ {\sigma\_{\mathcal{A}}}(a) \neq \varnothing $ for every $ a \in \mathcal{A} $. All proofs that I have seen of this result use the Hahn-Banach Theorem in one way or another (a typical proof may be found in Walter Rudin's *Real and Complex Analysis*). Hence, a natural question to ask would be: Can we remove the dependence of this result on the Hahn-Banach Theorem? Is it a consequence of ZF only? Otherwise, if it is equivalent to some weak variant of the Axiom of Choice (possibly weaker than the Hahn-Banach Theorem itself), has anyone managed to construct a model of ZF containing a Banach algebra that has an element with empty spectrum?



Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem.
I think Hahn-Banach can be eliminated from the usual proof, but
being a non-expert in set theory, I cannot guarantee that the proof
is completely independent of the axiom of choice.


Here is a sketch of a basic calculus proof. A function $U\to B$ 
from a region $U\subset C$ to a Banach space $B$ is called analytic if
every point has a neighborhood where it is represented by a
convergent Taylor series.
You can prove a weak form of Cauchy theorem which says that if a function
is analytic in 
$| z | < R \leq \infty$
 then its Taylor series
has radius of convergence at least $R$. It seems that this does not
use the axiom of choice.
Then you prove that Cauchy inequalities hold (there is a simple algebraic
proof of this, see my answer to [Liouville's theorem with your bare hands](https://mathoverflow.net/questions/116896/liouvilles-theorem-with-your-bare-hands/116910#116910)), and derive the Liouville theorem for Banach-space-valued
functions.


Then again it is an elementary fact that if $a-\lambda\_0 1$ has
has a bounded inverse then the resolvent is an analytic function (in the sense I defined above) in
a neighborhood of $\lambda\_0$. Then you show that if the resolvent exists everywhere
then it tends to $0$ as $\lambda\to\infty$. Then it seems to me that you obtain a proof
without Hahn-Banach by applying the Liouville theorem to the resolvent.


Sorry if I missed something...


EDIT. The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved, see
[Liouville's theorem with your bare hands](https://mathoverflow.net/questions/116896/liouvilles-theorem-with-your-bare-hands/116910#116910).


