maximal tree by Zorn's lemma In using the Zorn's lemma to show that every connected graph contains a spanning tree, we let $\{T_{\lambda}: \lambda \in \Lambda \}$ be a family of trees contained in $X$ which is totally ordered by inclusion. But how to show that $\bigcup\limits_{\lambda \in \Lambda} T_\lambda$ is a also a tree in $X$?
 A: Actually, $\bigcup\limits_{\lambda \in \Lambda} T_{\lambda}$ may not be a tree. 
You have to show that every chain in your family of trees has an upper bound. So, given a chain $\{T_{\alpha}$| $\alpha \in A$}, take the union $T=\bigcup\limits_{\alpha \in A} T_{\alpha}$. It is straightforward to show that this is tree: you just have to show that, given any two vertices $v_1,v_2 \in T$, there is exactly one path between them. 
(proof by contradiction: assume two paths - both the paths will be contained in some $T_{\alpha_0}$, but $T_{\alpha_0}$ is a tree, so you get a contradiction )
Now, since you have a poset where every chain has an upper bound, Zorn's lemma can be applied. 
A: The argument here is one of the common arguments when using Zorn's lemma.
If $G$ is a graph which is not a tree then it has a finite subgraph which witnesses that.


*

*If the chain was finite, then it has a maximal element. Therefore the union of the chain is that maximal element, and we are done. In fact we don't even care that the chain was finite, just that it had a maximal element.

*If the chain doesn't have a maximal element, and the union is not a tree then there is a finite subset which witnesses that. But that finite subset must have been added somewhere along the way. This counterexample, if so appears in some $T_\lambda$ in our chain. But we assumed that all the $T_\lambda$ are trees, so that is impossible.
Note that this is the same argument that the increasing union of linearly independent sets is linearly independent; that the increasing union of filters is a filter; and that the increasing union of ideals is an ideal, etc. The key is if the union wasn't a such object, it would mean that somewhere along the union we added a counterexample -- which contradicts our assumptions.
