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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.

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  • $\begingroup$ I think the key here is that he is fixing the bundle $x$. Notice the hyperplane is a function of which agent it is $i$ and what bundle he wants $x$. Might be helpful to stick to your $\mathcal{R}^2$ example and use $x = (1, 1)$ and $i$ gives us $b_i = B$. This should give you something like $H(i, x) = \{ p | p_1 = B - p_2 \}$ (where notationally I'm writing $p = (p_1, p_2) \in \mathcal{R}^2$). Does that help? $\endgroup$ – cc7768 Jun 3 '15 at 21:59
  • $\begingroup$ That does help a little - thanks. So in $\mathcal{R}^2$, a hyperplane is a line, though, right? And then for it to exactly equal the budget that line must go through $x$. But then, why can't I pivot that line around $x$ by a few degrees and call it a new hyperplane? What he describes as a hyperplane, $H(i,x)$, seems more like a set of hyperplanes to me. What am I missing? $\endgroup$ – Shane Jun 3 '15 at 22:06
  • $\begingroup$ Exactly, a hyperplane in $\mathcal{R}^2$ is a line. Are you thinking about the hyperplane being in price space or goods space? $x$ isn't in the same space as the hyperplane, so I don't know what you mean pivot the line around $x$. $\endgroup$ – cc7768 Jun 3 '15 at 22:09
  • $\begingroup$ I'm thinking of a space with two goods where $x$ is the $(1,1)$ coordinate. I'm thinking of the budget constraint hyperplane/line as being any line that goes through that point. Basically, I'm thinking of the standard intro micro plot with apples, oranges, and a budget constraint. I realize the budget set is just the vertices, not the hull, because the goods are indivisible. $\endgroup$ – Shane Jun 3 '15 at 22:12
  • $\begingroup$ This hyperplane is in the price space which means the axis that it would be plotted on are $p_1$ and $p_2$ so you can't really plot $x$ with this hyperplane $\endgroup$ – cc7768 Jun 3 '15 at 22:15
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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.

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