**Background:**

The complete infinite binary tree (CIBT) has path cardinality of continuum size, where path cardinality refers to the size of the set of all paths from the root.

If we consider the random infinite binary tree with constant degree $1+\epsilon$, for an infinitesimal constant $\epsilon$, we get a partial binary tree, which again has path cardinality of continuum size. We can see this by using a pull-down procedure, cutting out any long non-branching sub-path and reconnecting the ends, to show that this sparsely branching tree is isomorphic with the CIBT (see here: Path cardinality for random $(a+b)$-ary infinite trees). It is tempting to believe that all random binary trees, i.e. including those of completely random variant degree almost surely above one, can be pulled down in this manner to become isomorphic with CIBT.

*Sub-question: can all partial infinite binary trees of continuum path cardinality be pulled-down, so at least a subset can be put in isomorphic correspondence with the complete infinite binary tree?*

Consider now a mapping of CIBT to the $(x,y)$ plane, wherein, the nodes, $(a_i,b_i)$, are as follows (read from root up):

....

Level-4: $(-2-\frac{1}{2}- \frac{1}{4}- \frac{1}{8}, 1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8})$ ……..

Level-3: $(-2-\frac{1}{2}- \frac{1}{4}, 1+ \frac{1}{2}+ \frac{1}{4})$ ……..

Level-2: $(-2- \frac{1}{2},1+ \frac{1}{2})$ $(-2+ \frac{1}{2}, 1+ \frac{1}{2})$ $(2- \frac{1}{2},1+ \frac{1}{2}) (2+ \frac{1}{2}, 1+ \frac{1}{2})$

Level-1: $(-2,1) (2,1)$

root: $(0,0)$

Where the nodes are connected in order, with the usual edges of a binary tree.

Since $\sum_{i=0}^{\infty} \frac{1}{2^i} = 2$, we have now mapped the CBIT into the $(x,y)$ plane so that it forms a visual tree, with non-colliding branches, that extend uniformly from $(0,0)$ upwards, becoming increasingly dense as it approaches its limit, the number line spanning from $(-2,2)$ to $(2,2)$.

Let us call this number line the boundary of CIBT.

Since each point on the boundary corresponds to exactly one infinite path in CIBT, we can conclude that the boundary of CIBT is of continuum size, and is indeed an open real number line of length 4. We could of course instead transform to a finite height, infinitely spanning tree with a boundary covering the entire range of the real number line.

*Sub-question: will any binary tree of continuum path cardinality have a boundary containing at least one open subset of dimension 1?*

Consider now a procedure to collapse a tree from the top, wherein a node collapses, i.e. is removed from the tree along with its connected edges, if both its children nodes have already collapsed. We define the collapse of a tree, as the recursive collapse that happens once a set of elected leaf vertices are collapsed.

**Question:**

Let us select a set, F, as a fractal one-dimensional set, uniformly dense, with a Hausdorff dimension strictly less than 1. Let us for example set F to be of dimension equal to the inverse of a googol (i.e. pretty close to zero)

We now impose F on the boundary of CIBT, and collapse CIBT with F, obtaining the F-collapse of CBIT, which we can call FIBT.

We have derived FIBT, a binary tree with the boundary F.

*Sub-question: Is FIBT well-defined and unique? (even if informally stated here)*

The main question is now: What is the path cardinality of FIBT?

**Speculations:**

Obviously, FIBT has path cardinality of at most Continuum size.

Clearly, FIBT has path cardinality larger than countably infinite, as it has at least the cardinality of a set of positive Hausdorff dimension.

Now, let us consider whether we can use the pull-down procedure to show that it can be put in one to one correspondence with CIBT.

We can first observe that the infinite random tree of degree 1+e has the same boundary as CIBT, a boundary of uniform dimension 1. “Viewed from “the root” it is very sparse, but it becomes very dense at the boundary. FIBT has the reverse “appearance”, dense at the root, and increasingly sparse at the boundary.

A topological argument can proceed by viewing FIBT as an elastic structure, with nodes, connected by rubber bands, which we can cut and reconnect. Firstly, it is clear that cutting out long non-branching sub-paths and reconnecting at each end (basic pull-down) will not make any difference at the boundary. Secondly, we can then hope to cut and move infinite sub-trees, reconnecting them at other vertices, in order to show that a resulting part of the transformed FIBT achieves a unitary dimension at the boundary. However, an infinite number of additions of subsets, of the same dimension, will result in a set of the same dimension, so there seems to be no hope that any process of cutting and reconnecting can produce a set with different behavior at the boundary.

Since the pull-down procedure or any infinite cut-and-reconnect process cannot put FIBT in one-to-one correspondence with CIBT, it is tempting to think that FIBT has below continuum-size path cardinality.