I am quite new to discrete mathematics. I was going through graph and found this expression which was for full binary tree, where $B_h$ represents graph i.e. full binary tree of length/ height h $B_h := ([2]^{≤h}, \{\{u,ux\} \ such\ that\ u\epsilon [2]^{<h}, x\epsilon [2]\})$
I got confused here. what do $[2]^{≤h}$ and $[2]^{<h}$ mean here, i am quite certain that [2] means {1,2}. But that exponent on the set is quite not clear to me. does that mean cartesian product? I would be thankful for any kind of help. thanks!
If you will permit, I am more comfortable thinking of $[2]$ as $\{0,1\}$.
$[2]^h$ is the Cartesian product of $[2]$ with itself $[h]$ times: $$ [2]^h=[2]\times[2]\times\dots\times[2]\cong\{(b_1,b_2,\dots,b_h)\mid b_i\in [2]\text{ for }1\le i\le h\} $$
In other words, $[2]^h$ consists of binary sequences of length $h$. Then, $$ [2]^{\le h}=[2]^0\cup [2]^1\cup \dots\cup [2]^h $$ is the set of binary sequences of length at most $h$. That is, $2^{\le [h]}$ is not a Cartesian product, but a union of Cartesian products. You can probably imagine now what $[2]^{<h}$ is; it consists of binary sequences whose lengths are strictly less than $h$.
The set of edges is of the form $\{u,ux\}$, where $u$ is a binary sequence of length less than $h$, and $x\in \{0,1\}$. The multiplication here is concatenation. For example, letting $u=(0,1,0)$, and $x=0$, there is an edge between $(0,1,0)$ and $(0,1,0,0)$; we add a $0$ to the end of $u$ to get its neighbor.