Here is Zorn's lemma:
(Zorn) Every partially ordered set containing upper bounds on every linearly ordered subset contains a maximal element. (equivalent to the axiom of choice, proof omitted)
In my confusion I think I came up with a proof of Konig's lemma, different than the one that is normally given… I'm honestly just trying to make sure that it works.
(Konig) Every $\omega$-tree has an $\omega$-branch.
PROOF. Let $(T, \leq)$ be an $\omega$-tree, we show every linearly ordered collection of branches contains an upper bound with respect to $\subseteq$. Suppose we have a collection $\{S_\beta\}_{\beta<\alpha}$ of branches ordered linearly by $\subseteq$. Consider $S^*=\bigcup_{\beta<\alpha} S_\beta$, clearly this is a branch, i.e. a linearly ordered (with respect to $\leq$) subset of $T$ such that for all $t \in S^*$ if $s<t$ then $s \in S^*$. For every branch $S_\beta$ we have $(S_\beta, \leq) \preceq (S^*, \leq)$, so $S^*$ forms an upper bound for $\{S_\beta\}_{\beta<\alpha}$. Since every linearly ordered collection of branches is bounded above by a branch, and since $(T, \leq)$ is an $\omega$-tree, by Zorns lemma there is a maximal branch $(S, \leq)$ of order type $\omega$.
Does that work?
Best Answer
No. Just because the tree is an $\omega$-tree it does not mean there are no maximal nodes. And since you have no control over this, you don't know if your maximal branch is just finite, below one of the maximal nodes in the tree.
What you need to do first is prove that every $\omega$-tree can be pruned. Namely, we can remove all the maximal nodes, then repeat the process, and at the end if we have removed all the nodes in the tree have a contradiction.
Once the tree is pruned, your argument will work.