0,1-Tree, number of branches

Question

How many branches has this tree: $\{s\in 2^{<\kappa}:|\alpha\in dom(s):s(\alpha)\neq 0|<\aleph_0\}$, and why? How does the tree look like, i.e. what is its shape? Thank you.

Answer 1

Well, this tree will look sort of like the full binary tree $2^{<\kappa}$, except that all of the nodes with infinitely many $1$s are missing. Let's call your tree $T$. Then every branch of $T$ can be extended to a branch of $2^{<\kappa}$, i.e. a function $\kappa\to 2$. But given a function $\kappa\to 2$, not all of its restrictions will be elements of $T$: only those with only finitely many $1$s. That is, the restriction forgets about all of the values of the function $\kappa\to 2$ after the first $\omega$ $1$s.

Thus the branches of $T$ are in bijection with subsets $S$ of $\kappa$ of order type $\leq\omega$. Explicitly, given such a subset $S$, you get a branch by taking all initial segments of the characteristic function of $S$ prior to the supremum of its first $\omega$ elements. Conversely, given a branch $B$, the union of all the elements of $B$ is a function $f:\alpha\to 2$ for some $\alpha\leq\kappa$. If $\alpha<\kappa$, the fact that $B$ cannot be extended by adding $f$ as another member means that $f$ takes the value $1$, which means $f$ is the characteristic function of a cofinal subset of $\alpha$ of order-type $\omega$. If $\alpha=\kappa$, then $f$ must similarly be either the characteristic function of a finite subset of $\kappa$ or the characteristic function of a cofinal subset of $\kappa$ of order-type $\omega$. Combining all the cases together, $f$ comes from a subset of $\kappa$ of order-type $\leq\omega$.

Let us now count how many branches there are (assuming $\kappa$ is an infinite cardinal). There are $\kappa$ finite subsets of $\kappa$. The elements of a subset of order-type $\omega$ can be chosen one-by-one in increasing order, and at each step you have $\kappa$ choices (there are still $\kappa$ elements greater than the last one you chose). So there are $\kappa^{\aleph_0}$ subsets of order-type $\omega$. In total, then, there are $\kappa+\kappa^{\aleph_0}=\kappa^{\aleph_0}$ branches.