Given a perfect binary tree which at a certain fixed level has infinitely many leaves, hence each leaf at that fixed level has infinitely many ancestors. From that fixed level the tree also extends to infinite height.
Are the nodes of this tree countably infinite or uncountably infinite?
I know that a rooted binary tree, even if extended to a height of infinity is countably infinite; just start at the root and enumerate level by level. However, I cannot imagine a way to count the nodes of this tree or to apply the idea that the union of countably infinite subtrees is countable. Thus, I’m leaning towards it being uncountably infinite but have serious doubts.
How can a tree look like that?
Imagine that instead of the tree simply growing in height it is also expanding sideways as shown below, so it is infinitely high, infinitely wide and infinitely deep. Getting infinitely wide is illustrated below.
Best Answer
It sounds like you're considering an iterative construction which keeps embedding a binary tree inside a larger binary tree, always growing a bit in each direction. If so, this actually does stay countable: picking an arbitrary node and "rotating" everything around that node, what you have is just three copies of the usual (one-way-)infinite binary tree connected at that chosen node.
Of course it's a bit difficult to prove this, given that the definition you have is still rather vague. But I can't see a way of interpreting what you have in mind which doesn't lead to a countable graph. Certainly any time I define a sequence of finite graphs $G_1\subseteq G_2\subseteq G_3\subseteq ...$, their union $$\bigcup_{i\in\mathbb{N}}G_i$$ is countable, so if you want to wind up with something uncountable your construction has to be substantially more complicated.