The answers here are quite mathematical. I hope somebody can explain this particular point here.
Why is it for a tree with height $\omega$ that all its levels need to be finite in order to have an infinite branch ? This question is about the necessity of the finiteness of the levels. For example what is false about the picture below ? I don't see why existance or the absence of the (encircled) subset on the right, affects the existance of the infinite branch ?
Definitions
A tree $(T,<)$ is a partially ordered set with $\forall y \in T:\ T_{<x} := \{y\in T|y<x\}\ \text{is well-ordered}$.
The height of an element $x$ is the ordinal $ht(x,T):=\alpha_x \cong T_{<x}$, i.e. of the order-type of $T_{<y}$.
The $\alpha$th level of the tree is $T(\alpha) = \{x\in T|\ ht(x,T) \cong \alpha\}$
A branch is a subset of $T$ that is maximal chain
The height of the tree is $ht(T) = \sup\{ht(x,T)+1|\ x \in T\}$,
Added:
I doubt about this last definition, for if we have an $\omega$-long branch, the tree would be $(\omega+1)$-heigh. Though this definition is found also in Kunen, Set Theory, An Introduction to Independence Proofs (1992), §5 Trees, p. 68. Could anyone explain what is wrong with this ?
Best Answer
Take a collection of finite trees, such that your collection contains trees of every possible finite height.
Graft them all onto a single new root node.
Then the combined tree has height $\omega$ (it certainly can't have any finite height) -- but it can't contain any infinite branch. Such a branch would have to contain one of the successors of the new root node. But that successor is the root of one of the original trees, so it cannot be in an infinite branch.