Budget constraint hyperplanes This question relates to a specific paper by Eric Budish, published in 2011 in JPE, but I've tried to put all relevant information in this question. On page 1072, he defines budget constraint hyperplanes as follows:


  Let $H(i,x) = \{\mathbf{p}:\mathbf{p} \cdot x = b_i \}$ denote the
  hyperplane in $M$-dimensional price space along which agent $i$ can
  exactly afford bundle $x$. As prices cross $H(i,x)$ from below, bundle
  $x$ goes from being affordable for $i$ to being unaffordable for $i$.


$\textbf{p}$ represents a price vector for the $M$ goods, which, importantly, are indivisible. $b_i$ is agent $i$'s budget. I think the rest is sufficiently self-explanatory. My confusion is with this subsequent statement:


  Importantly, the number of such hyperplanes is finite because the
  number of agents and the number of bundles are finite. This is an
  advantage of having only indivisible goods.


This I don't see. For example, suppose $M=2$. Then aren't all $\textbf{p}=(\alpha b_i, (1-\alpha)b_i)$ such that $\alpha \in [0,1]$ hyperplanes meeting that definition, and thus I have infinitely many?

Having said that, a hyperplane is the set of those price vectors, not each of the vectors, so maybe the full set of price vectors I've just described together define just one hyperplane? I suppose my issue is understanding what a hyperplane is in this setting, and how the indivisibility gives us just a finite set of hyperplanes to work with. Any guidance would be very much appreciated.
The set $\{\mathbb{p}:\exists \alpha \in (0,1), \mathbb{p}_1=\alpha b_i, \mathbb{p}_2=(1-\alpha)b_i\}$ coincides with the hyperplane $\{\mathbb{p}:p_1+p_2=b_i\}$, therefore you are not finding extra hyperplanes by doing this and varying $\alpha$.

More generally, if you fix $x=(x_1,x_2)$ and $b_i$, you obtain one hyperplane characterized by the equation $\{\mathbb{p}:p_1 x_1+p_2 x_2=b_i\}$. Since there are finitely many such $x$ (and $b_i$ is given), there are finitely many such hyperplanes.