# Aggregation of the closure property of a production set

Consider an economy with finitely many goods $$I$$ and factors $$F$$. For each good $$i$$, let $$Y_i \subset \mathbb R^I \times \mathbb R^F_{\geq 0}$$ denote a production set for $$i$$. Assume each $$Y_i$$ has constant returns to scale and for all $$(y,l)\in Y_i$$, $$y_{j \neq i} \leq 0$$. Finally, denote by $$Y \equiv \{ (\sum_{i \in I}y^i,\sum_{i \in I}l^i) \ | \ \forall i \in I, (y^i,l^i) \in Y_i\}$$ the canonical aggregate production set.

It is common in GE theory to assume that $$Y$$ is closed.

Is it known what conditions on the sets $$Y_i$$ guarantee this?

To see that it is not sufficient to simply assume each $$Y_i$$ is closed, consider the following counterexample: There are two goods and one factor. Given one unit of labor, each firm can produce $$n + \frac{n}{n+1}$$ units of output per $$n$$ inputs of the other good used. By taking a limit as $$n\to \infty$$ for each firm, one may see that the closure of $$Y$$ contains $$(y,l) = ((1,1),2)$$

• Your counterexample seems to violate constant returns to scale. So do we assume that or not? Jun 3 at 4:36
• An example that satisfies CRS can be found on page 4 here, though it is not written in economic language. Jun 3 at 8:47

Doing this more abstractly, let $$Y_j\subseteq\mathbb{R}^n$$ be a production set for $$j=1,\ldots,m$$ and let $$Y=Y_1+Y_2+\cdots+Y_N=\{y_1+y_2+\cdots+y_n|y_j\in Y_j, j=1,\ldots,m\}$$ be the aggregate production set. The standard result on when the aggregate production set is closed is the following:

Theorem: Let $$Y_j$$ be closed and convex sets containing $$0$$ for $$j=1,\ldots,m$$ and assume that $$Y\cap -Y=\{0\}$$. Then $$Y$$ is closed.

Now, this result uses an assumption (irreversibility and possibility of inaction) on the aggregate production set, but this follows from $$0$$ being in every production set and the input and output commodities being separate in your setting. The assumption that $$0$$ is in each individual production set is not needed, but the proof becomes messier (though not fundamentally different) without it.

The answer of tdm points you to a proof via asymptotic cones. Here is a more direct argument:

Proof: Let $$(y_m)\to y$$ be a convergent sequence in $$Y$$. For every $$m$$, let $$(y^1_m,\ldots,y^n_m)\in Y_1\times\ldots\times Y_n$$ be such that $$y_m=\sum_{j=1}^n y^j_m$$. If this sequence is bounded, a compactness argument guarantees that $$y\in Y$$. We show that the sequence has to be bounded. Suppose not. W.l.o.g., assume that none of the terms in this sequence is zero and that the norm of the terms is increasing and unbounded. Define a sequence $$(x^1_m,\ldots,x^n_m)$$ by $$x^j_m=y^j_m/\|(y^1_m,\ldots,y^n_m)\|$$. The sequence has a converging subsequence $$(z^1_m,\ldots,z^n_m)\to(z^1,\ldots,z^n)$$ because it lies in the compact unit-sphere and $$(z_1,\ldots,z^n)\in Y_1,\ldots, Y_n$$ because each $$Y_j$$ is convex and contains $$0$$. Also, $$\|(z^1,\ldots,z^n)\|=1$$, so at least one coordinate must be nonzero. W.l.o.g. let $$z^1\neq 0$$. We have $$\sum_{j=1}^n z^j=\lim_{m\to\infty} \frac{y_m}{\|(y_m^1,\ldots,y_m^n)|\|}=0$$ because $$\|(y^1_m,\ldots,y^n_m)\|$$ is unbounded and the convergent sequence $$(y_m)$$ is bounded. So $$z^1-\sum_{j=2}^n z^j=0$$, but $$z_1\in Y$$ and $$\sum_{j=2}^n z^j\in Y$$ because each $$Y_j$$ contains $$0$$, so this contradicts $$Y\cap-Y=\{0\}$$.

• When you say (in the proof) "If this sequence is bounded...." do you mean "If the sequences $(y^j_m)$ are bounded...". I guess $(y_m)$ is bounded by assumption (as it converges). Or do you mean the sequence $(y^1_m, \ldots, y^n_m)$ as a sequence in $Y_1\times \ldots \times Y_n$?
– tdm
Jun 3 at 8:56
• Yes, I mean the latter sequence in the product space. Jun 3 at 9:05
• Thanks Michael, this is excellent. Also I suppose the convexity assumption is not necessary in my setting since I am happy to assume CRS. Do you know of a citation for this argument? Thanks again. Jun 3 at 14:13
• @JohnSturm I guess the argument is similar to what one proves when one proves the relevant results about asymptotic cones and recession cones. But the argument comes from my teacher K. Podczeck, who tried to come up with his own proof for the result in Debreu's book. Jun 3 at 18:19
• Thank you! Much appreciated. Jun 10 at 1:08

For a fully overview on the conditions for the sum of closed sets to be closed, see this note of Kim Border

### Recession cones

I'll be working with subsets of $$\mathbb{R}^n$$. Let's start with some definitions.

Def: A set $$C$$ is convex if for $$x, y \in C$$ and $$\alpha \in [0,1]$$, $$\alpha x + (1-\alpha) y \in C$$.

Def: A set $$K$$ is a cone if for $$x \in K$$ and $$\lambda \ge 0$$, $$\lambda x \in K$$.

Def: The recession cone $$0^+C$$ of a set $$C$$ is defined as: $$0^+ C = \{z \in \mathbb{R}^n| \forall x \in C, \forall \alpha \ge 0, x + \alpha z \in C\}.$$

The idea behind the recession cone is that it captures all directions of unboundedness of $$C$$. For example if $$(1,1) \in 0^+C$$ this means that $$C$$ is unbounded in the direction $$(1,1)$$: for all $$x \in C$$ and all $$\alpha \ge 0$$, $$x + \alpha\cdot (1,1) \in C$$.

Here $$x + \alpha(1,1)$$ is an infinite ray (half line) with slope 1 starting at $$x$$. The element $$(1,1)$$ being in $$0^+C$$ requires that all these half lines (over all $$x \in C$$) are in $$C$$.

Now take two sets $$C_1$$ and $$C_2$$ that are closed. We would like to have a conditions such that $$C_1 + C_2$$ is also closed. In order to do this, we need that for all converging sequences $$z_n \to z$$ in $$C_1 + C_2$$ we can find a convergent sequence $$x_n \to x$$ in $$C_1$$ and a convergent sequence $$y_n \to y$$ in $$C_2$$ such that $$x_n + y_n = z_n$$.

One thing that could go wrong when checking this condition is that, although $$z_n = x_n + y_n$$ converges both $$x_n$$ and $$y_n$$ go towards infinity in opposite directions, so they are not converging. This could happen if the recession cones $$0^+ C_1$$ and $$0^+ C_2$$ contain non-zero vectors that add up to zero.

For example, if $$(1,1) \in O^+ C_1$$ and $$(-1, -1) \in 0^+ C_2$$. Then for som $$x \in C_1$$ and $$y \in C_2$$ we can define: $$z_n = \underset{x_n}{\underbrace{x + n(1,1)}} + \underset{y_n}{\underbrace{y + n(-1,-1)}} = x + y.$$ Here $$z_n = x_n + y_n$$ converges although both $$x_n \equiv x + n(1,1)$$ and $$y_n \equiv y + n(-1,-1)$$ diverge.

The following definition captures this idea.

Def: let $$K_1, \ldots, K_n$$ be cones. We say that they are positively semi-independent if for all $$x_i \in K_i$$ $$x_1 + x_2 + \ldots + x_n = 0 \to x_1 = x_2 = \ldots = x_n = 0.$$

The following theorem (which proof I will skip, see the note of Border) shows that for closed convex sets, positive semi-independence of the recession cones is indeed a sufficient condition.

Th Let $$C_1,\ldots, C_n$$ be a collection of closed and non-empty convex sets. If the recession cones $$0^+C_1, 0^+ C_2, \ldots 0^+ C_n$$ are positively semi-independent then $$\sum_i C_i$$ is closed.

### CRS production possibility sets

Let's go back to the question. Let $$Y$$ be the production possibility set of a constant returns to scale production function. CRS is equivalent to the requirement that $$Y$$ is a cone. In particular: if $$y \in Y$$ and $$\alpha \ge 0$$ then $$\alpha y \in Y$$.

The following holds for convex cones:

Th: If $$Y$$ is a non-empty convex cone, then $$0^+ Y = Y$$.

Proof As $$Y$$ is a cone, we have that $$0 \in Y$$. Now, let $$z \in 0^+ Y$$. Then for all $$x \in Y$$ and all $$\alpha \ge 0$$: $$x + \alpha z \in Y$$. Setting $$x = 0$$ and $$\alpha = 1$$ gives $$z \in Y$$. For the reverse, let $$z \in Y$$, $$\alpha \ge 0$$ and $$x \in Y$$. As $$Y$$ is a cone, we have $$\alpha z \in Y$$, and as $$Y$$ is a convex cone, we have $$2(\frac{1}{2} x + \frac{1}{2}\alpha z) = x + \alpha z \in Y$$. As $$\alpha \ge 0$$ and $$x$$ were arbitrarily, this shows that $$z \in 0^+ Y$$.

We then have the following Corollary whose proof follows from the previous two theorems.

Corr: Let $$Y_1,\ldots, Y_n$$ be a set of closed, convex and CRS production possibility sets. If $$Y_1, \ldots, Y_n$$ are positively semi-independent, then $$Y = \sum_i Y_i$$ is closed.

Towards the interpretation of this condition. Assume for a moment that we only have two closed and convex CRS production possibility sets $$Y_1, Y_2$$ and assume that they are not positively semi-independent. This means that we can find $$x \in Y_1$$ and $$y \in Y_2$$ such that $$x \ne 0$$ and $$x + y = 0.$$ This means that $$x \in Y_1$$ and $$y = -x = Y_2$$. In other words, we can use the first technology to produce $$x$$ and then, in turn, the second technology to produce $$-x$$. Intuitively, this corresponds to a reversible technology. For example, you use, labour, wood and nails to make a table ($$x$$) and then, you can use this table as an inputs to make back the original inputs (labour, wood and nails) ($$-x$$). In physics, this is called a perpetual motion machine (which violates the first or second law of theromodynamics). So if all inputs are taken into account, the closed, convex, CRS production possibility sets should indeed by positively semi-independent and their sum should be closed.

• Thanks for this thorough answer! Jun 3 at 15:00