The cardinality of the set of all root paths in the infinite complete binary tree is equal to the cardinality of the Continuum. The same holds true for k-ary trees for any finite k. But what is the case for k infinite?
[Math] Cardinality of the set of all paths in the infinite complete infinitary tree
set-theory
Related Solutions
Your title question seems to be answered by the following theorem:
Theorem. The set of infinite paths through any finitely branching tree is either countable or size continuum.
Proof. This is a consequence of the Cantor-Bendixson theorem. To summarize, suppose that $T$ is a finitely branching tree. Since we are only interested in the set of paths through $T$, we may assume without loss by pruning that there are no leaves in $T$. A branch $b$ through $T$ is isolated, if above some level in the tree, there is no branching off of $b$. One now performs the transfinite process of computing Cantor-Bendixson derivatives, where at each stage, we cast out the isolated points. Specifically, let $T_0=T$ be the initial tree, and let $T_{\alpha+1}$ be the tree obtained by cutting off the isolated points of the tree $T_\alpha$ (and pruning any leaves that may be left as residue). At limit stages, we take $T_\lambda=\bigcap_{\alpha\lt\lambda} T_\alpha$. The Cantor-Bendixson analysis is that there is some countable ordinal $\alpha$ for which $T_\alpha$ is either empty or has no isolated branches, and this ordinal is called the Cantor-Bendixson rank of the tree (or of the set of paths through the tree). If a nonempty tree has no isolated branches, then the set of branches through it is a perfect set, a closed set with no isolated points. Thus, the original set of branches is the union of a countable set with a perfect set, namely, the countably many branches that became isolated along the way and the perfect set of branches left at the end. Every nonempty perfect set is easily seen to have size continuum, and so the set of paths is either countable or has size continuum. QED
So if you've got a tree, then either there are only countably many paths, in which case the fractal dimension will be zero, or there will be continuum many paths.
Original answer:
This was too long for a comment.
Sub-question: can all partial infinite binary trees of continuum path cardinality be pulled-down, to be put in isomorphic correspondence with the complete binary tree?
No, because a binary tree can have isolated points, even if it has continuum many branches, but the complete binary tree has no isolated points. Imagine a tree with a lonely branch extending to the left, but a fully branching component (and hence continuum many paths) coming out of a node to the right.
Sub-question: will any binary tree of continuum path cardinality have a boundary containing at least one open subset of dimension 1?
No, because the Cantor set is the boundary of the associated binary tree, but the Cantor set is nowhere dense. For example, in your tree, take the subtree determined by successively going always left plus left again or right plus right again. The corresponding set of continuum many paths contains no open set.
The main question is now: What is the cardinality of FIBT?
Your question is problematic. The set $F$ is determined by a subtree only when $F$ is a closed set, since the set of paths through a tree is always closed. I don't know what you mean by "uniformly dense", but natural measure-theoretic interpretations of this are impossible by the Lebesgue density theorem. (If you are asking about the cardinality of the tree whose boundary is a closed set $F$, then it is countable, since the whole tree has only countably many nodes.) You indicate in your edit that you are asking for the cardinality of the set of paths through the tree. But the cardinality of the set of paths through any partial binary tree is always either countable or size continuum. This corresponds to the fact proved by Cantor that every closed set is either countable or size continuum. In other words, Cantor proved that the continuum hypothesis holds in the case of closed sets.
It is consistent that both of the questions have a negative answer. Indeed, this happens if MA holds.
A set $E$ of reals is called a Luzin set if $E$ has size continuum and for every meager set $X$ the intersection $E\cap X$ has size less than continuum.
A set of reals $E$ is called a Sierpiński set if $E$ has size continuum and for every measure zero set $X$ the intersection $E\cap X$ has size less than continuum.
Theorem: MA implies that there are Luzin and Sierpiński sets.
Proof: To construct a Luzin set, list all Borel nowhere dense sets in order type continuum: $\langle F_\alpha;\alpha<\mathfrak{c}\rangle$. For each $\alpha$ choose some $e_\alpha\notin \bigcup_{\beta<\alpha}F_\beta$; this is possible since MA implies that the union of less than continuum meager sets is meager. $E=\{e_\alpha;\alpha<\mathfrak{c}\}$ has size continuum and its intersection with every closed nowhere dense set has size less than continuum by construction. But since a meager set is contained in a union of countably many closed nowhere dense sets, $E$ must be a Luzin set.
To construct a Sierpiński set, replace "Borel nowhere dense" above by "Borel of measure zero" and "meager" by "measure zero". $\square$
In particular, this shows that assuming cardinal characteristics are large is not helpful for this problem.
The same avoidance idea seems to also show:
Theorem: If $V$ was obtained from $W$ by adding more than $\mathfrak{c}^W$ many Cohen (or random) reals to $W$, then the set of generic reals is Luzin (or Sierpiński) in $V$.
Best Answer
Assuming your path has countable length, the set of all paths in a $k$-ary tree will have cardinality $k^{\aleph_0}$. Indeed, at each step you have $k$ choices, and there are $\aleph_0$ steps (think of a path as a function from $\mathbb{N}$ to $[k]$).