I'm reading the following proof, where the closure of the rational polyhedron $P$ is denoted $P'$.
I don't get the line where $Y$ is defined. This is a set of linear expressions of the form $y^TA$, and since $y \in \mathbb{R}$, I don't suppose that the "boundedness" here is on the size of the set, but instead on the space it carves out in $\mathbb{R}^n$? But if so how do we define it without the right hand sides of $y^TA$?
$y$ is not supposed to be a real number, it is a vector.
A better notation would be
$$Y:=\{y^TA: y \in [0,1]^m\}$$
Since $A \in \mathbb{Z}^{m \times n}$ and $y^T \in \mathbb{R}^{m \times 1}$, we have $y^TA \in \mathbb{R}^{1\times n}$ and hence we have $Y \subseteq \mathbb{R}^n$, it is bounded since it is a continuous image of a compact set. By boundedness of a set $S$, it means we can find $r>0$, such that for a distance function $d$, for all $u, v \in S$, we have $d(u,v) \le r$.