I'm looking to use Zorn's lemma on the follwoing set:
Let $\Sigma$ be the set of all pairs $(A,f)$ where $A$ is a subring of field $K$ and $f$ its homomorphism of $A$ into $\Omega$. We partially order the set as follows:
$$(A,f) \leq (A',f') \leftrightarrow A \subset A' \land f'|_{A} = f$$
Now, given a chain $\{(A_i,f_i)\}_{i \in \{1...n\}} \subset \Sigma$ I want to show it has an upper bound $(\bar{A}, \bar{f})$. I believe $\bar{A} = \cup_{i =1}^n A_i$ now I want to define $\bar{f}$ such that $\bar{f}|_{A_i} = f_i \; \; \forall i \in \{1...n\}$
Since its a chain I'm working with I think there's no harm assuming the functions all satisfy that $f_j|_{A_i} = f_i \; \forall \; i\leq j $ but I still arrive nowhere.
Any help would be apreciated, I'm not sure if I'm complicating things more than I should.
Defining $\overline f$ requires some care.
We define, for $a\in A=\bigcup A_i,$ $\overline f(a)=f_k(a)$ when $a\in A_k.$
Then we have to prove that this is well-defined - that if $a\in A_k$ and $a\in A_{j},$ then $f_k(a)=f_j(a).$
This isn’t hard - one of $(A_k,f_k)$ and $(A_j,f_j)$ is $\leq$ the other in the order, since they are in a chain.
A more set theoretic approach to this is to define $f\subset A\times \Omega$ as:
$$f=\{(a,\omega)\mid \exists k(a\in A_k\land f_k(a)=\omega)\}$$
What we have here is just a relation between $A$ and $\Omega.$ We need to prove it is a function. That is, for each $a\in A,$ there is exactly one $\omega\in \Omega$ so that $(a,\omega)\in f.$
This amounts to proving the same thing as proving $f$ was well-defined in the first section of this answer.
We usually skip this set theory work for that reason - we know what we are “really doing” when we write the first approach. The set theory adds noise to the argument.
Of course, once you’ve defined $\overline f,$ you need to prove it is a homomorphism. That isn’t hard, but it is a necessary step.