<p>Doing this more abstractly, let <span class="math-container">$Y_j\subseteq\mathbb{R}^n$</span> be a production set for <span class="math-container">$j=1,\ldots,J$</span> and let
<span class="math-container">$$Y=Y_1+Y_2+\cdots+Y_J=\{y_1+y_2+\cdots+y_J|y_j\in Y_j, j=1,\ldots,J\}$$</span>
be the aggregate production set. The standard result on when the aggregate production set is closed is the following:</p>
<p><strong>Theorem:</strong> Let <span class="math-container">$Y_j$</span> be closed and convex sets containing <span class="math-container">$0$</span> for <span class="math-container">$j=1,\ldots,J$</span> and assume that <span class="math-container">$Y\cap -Y=\{0\}$</span>. Then <span class="math-container">$Y$</span> is closed.</p>
<p>Now, this result uses an assumption (irreversibility and possibility of inaction) on the aggregate production set, but this follows from <span class="math-container">$0$</span> being in every production set and the input and output commodities being separate in your setting. The assumption that <span class="math-container">$0$</span> is in each individual production set is not needed, but the proof becomes messier (though not fundamentally different) without it.</p>
<p>The answer of tdm points you to a proof via asymptotic cones. Here is a more direct argument:</p>
<p><strong>Proof:</strong>
Let <span class="math-container">$(y_m)\to y$</span> be a convergent sequence in <span class="math-container">$Y$</span>. For every <span class="math-container">$m$</span>, let <span class="math-container">$(y^1_m,\ldots,y^J_m)\in Y_1\times\ldots\times Y_J$</span> be such that <span class="math-container">$y_m=\sum_{j=1}^J y^j_m$</span>. If this sequence is bounded, a compactness argument guarantees that <span class="math-container">$y\in Y$</span>. 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 <span class="math-container">$(x^1_m,\ldots,x^J_m)$</span> by <span class="math-container">$x^j_m=y^j_m/\|(y^1_m,\ldots,y^J_m)\|$</span>. The sequence has a converging subsequence <span class="math-container">$(z^1_m,\ldots,z^J_m)\to(z^1,\ldots,z^J)$</span> because it lies in the compact unit-sphere and <span class="math-container">$(z_1,\ldots,z^J)\in Y_1,\ldots, Y_J$</span> because each <span class="math-container">$Y_j$</span> is convex and contains <span class="math-container">$0$</span>. Also, <span class="math-container">$\|(z^1,\ldots,z^J)\|=1$</span>, so at least one coordinate must be nonzero. W.l.o.g. let <span class="math-container">$z^1\neq 0$</span>. We have <span class="math-container">$$\sum_{j=1}^J z^j=\lim_{m\to\infty}  \frac{y_m}{\|(y_m^1,\ldots,y_m^J)|\|}=0$$</span>
because <span class="math-container">$\|(y^1_m,\ldots,y^J_m)\|$</span> is unbounded and the convergent sequence <span class="math-container">$(y_m)$</span> is bounded.
So <span class="math-container">$z^1-\sum_{j=2}^J z^j=0$</span>, but <span class="math-container">$z_1\in Y$</span> and <span class="math-container">$\sum_{j=2}^J z^j\in Y$</span> because each <span class="math-container">$Y_j$</span> contains <span class="math-container">$0$</span>, so this contradicts <span class="math-container">$Y\cap-Y=\{0\}$</span>.</p>

CC BY-SA 4.0
2021-06-03T08:56:31.147Z
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
CC BY-SA 4.0
2021-06-03T09:05:12.653Z
Yes, I mean the latter sequence in the product space.
Michael Greinecker
CC BY-SA 4.0
2021-06-03T14:13:21.360Z
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.
John Sturm
CC BY-SA 4.0
2021-06-03T18:19:02.467Z
@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.
Michael Greinecker
CC BY-SA 4.0
2021-06-03T08:42:57.477Z
2023-10-21T14:05:08.060Z
2023-10-21T14:05:08.060Z
Michael Greinecker
<p>For a fully overview on the conditions for the sum of closed sets to be closed, see this note of <a href="http://www.its.caltech.edu/%7Ekcborder/Notes/AsymptoticCones.pdf" rel="nofollow noreferrer">Kim Border</a></p>
<h3>Recession cones</h3>
<p>I'll be working with subsets of <span class="math-container">$\mathbb{R}^n$</span>. Let's start with some definitions.</p>
<blockquote>
<p><strong>Def:</strong> A set <span class="math-container">$C$</span> is <strong>convex</strong> if for <span class="math-container">$x, y \in C$</span> and <span class="math-container">$\alpha \in [0,1]$</span>, <span class="math-container">$\alpha x + (1-\alpha) y \in C$</span>.</p>
</blockquote>
<blockquote>
<p><strong>Def:</strong> A set <span class="math-container">$K$</span> is a <strong>cone</strong> if for <span class="math-container">$x \in K$</span> and <span class="math-container">$\lambda \ge 0$</span>, <span class="math-container">$\lambda x \in K$</span>.</p>
</blockquote>
<blockquote>
<p><strong>Def:</strong> The <strong>recession</strong> cone <span class="math-container">$0^+C$</span> of a set <span class="math-container">$C$</span> is defined as:
<span class="math-container">$$
0^+ C = \{z \in \mathbb{R}^n| \forall x \in C, \forall \alpha \ge 0, x + \alpha z \in C\}.
$$</span></p>
</blockquote>
<p>The idea behind the recession cone is that it captures all directions of unboundedness of <span class="math-container">$C$</span>. For example if <span class="math-container">$(1,1) \in 0^+C$</span> this means that <span class="math-container">$C$</span> is unbounded in the direction <span class="math-container">$(1,1)$</span>: for all <span class="math-container">$x \in C$</span> and all <span class="math-container">$\alpha \ge 0$</span>, <span class="math-container">$x + \alpha\cdot (1,1) \in C$</span>.</p>
<p>Here <span class="math-container">$x + \alpha(1,1)$</span> is an infinite ray (half line) with slope 1 starting at <span class="math-container">$x$</span>. The element <span class="math-container">$(1,1)$</span> being in <span class="math-container">$0^+C$</span> requires that all these half lines (over all <span class="math-container">$x \in C$</span>) are in <span class="math-container">$C$</span>.</p>
<p>Now take two sets <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> that are closed. We would like to have a conditions such that <span class="math-container">$C_1 + C_2$</span> is also closed. In order to do this, we need that for all converging sequences <span class="math-container">$z_n \to z$</span> in <span class="math-container">$C_1 + C_2$</span> we can find a convergent sequence <span class="math-container">$x_n \to x$</span> in <span class="math-container">$C_1$</span> and a convergent sequence <span class="math-container">$y_n \to y$</span> in <span class="math-container">$C_2$</span> such that <span class="math-container">$x_n + y_n = z_n$</span>.</p>
<p>One thing that could go wrong when checking this condition is that, although <span class="math-container">$z_n = x_n + y_n$</span> converges both <span class="math-container">$x_n$</span> and <span class="math-container">$y_n$</span> go towards infinity in opposite directions, so they are not converging. This could happen if the recession cones <span class="math-container">$0^+ C_1$</span> and <span class="math-container">$0^+ C_2$</span> contain non-zero vectors that add up to zero.</p>
<p>For example, if  <span class="math-container">$(1,1) \in O^+ C_1$</span> and <span class="math-container">$(-1, -1) \in 0^+ C_2$</span>. Then for som <span class="math-container">$x \in C_1$</span> and <span class="math-container">$y \in C_2$</span> we can define:
<span class="math-container">$$
z_n = \underset{x_n}{\underbrace{x + n(1,1)}} + \underset{y_n}{\underbrace{y + n(-1,-1)}} = x + y.
$$</span>
Here <span class="math-container">$z_n = x_n + y_n$</span> converges although both <span class="math-container">$x_n \equiv x + n(1,1)$</span> and <span class="math-container">$y_n \equiv y + n(-1,-1)$</span> diverge.</p>
<p>The following definition captures this idea.</p>
<blockquote>
<p><strong>Def:</strong> let <span class="math-container">$K_1, \ldots, K_n$</span> be cones. We say that they are <strong>positively semi-independent</strong> if for all <span class="math-container">$x_i \in K_i$</span>
<span class="math-container">$$
x_1 + x_2 + \ldots + x_n = 0 \to x_1 = x_2 = \ldots = x_n = 0.
$$</span></p>
</blockquote>
<p>The following theorem (which proof I will skip, see the note of <a href="http://www.its.caltech.edu/%7Ekcborder/Notes/AsymptoticCones.pdf" rel="nofollow noreferrer">Border</a>) shows that for closed convex sets, positive semi-independence of the recession cones is indeed a sufficient condition.</p>
<blockquote>
<p><strong>Th</strong> Let <span class="math-container">$C_1,\ldots, C_n$</span> be a collection of closed and non-empty convex sets. If the recession cones <span class="math-container">$0^+C_1, 0^+ C_2, \ldots 0^+ C_n$</span> are positively semi-independent then <span class="math-container">$\sum_i C_i$</span> is closed.</p>
</blockquote>
<h3>CRS production possibility sets</h3>
<p>Let's go back to the question. Let <span class="math-container">$Y$</span> be the production possibility set of a constant returns to scale production function. CRS is equivalent to the requirement that <span class="math-container">$Y$</span> is a cone.  In particular: if <span class="math-container">$y \in Y$</span> and <span class="math-container">$\alpha \ge 0$</span> then <span class="math-container">$\alpha y \in Y$</span>.</p>
<p>The following holds for convex cones:</p>
<blockquote>
<p><strong>Th:</strong> If <span class="math-container">$Y$</span> is a non-empty convex cone, then <span class="math-container">$0^+ Y = Y$</span>.</p>
</blockquote>
<blockquote>
<p><em>Proof</em> As <span class="math-container">$Y$</span> is a cone, we have that <span class="math-container">$0 \in Y$</span>. Now, let <span class="math-container">$z \in 0^+ Y$</span>. Then for all <span class="math-container">$x \in Y$</span> and all <span class="math-container">$\alpha \ge 0$</span>: <span class="math-container">$x + \alpha z \in Y$</span>. Setting <span class="math-container">$x = 0$</span> and <span class="math-container">$\alpha = 1$</span> gives <span class="math-container">$z \in Y$</span>. For the reverse, let <span class="math-container">$z \in Y$</span>, <span class="math-container">$\alpha \ge 0$</span> and <span class="math-container">$x \in Y$</span>.  As <span class="math-container">$Y$</span> is a cone, we have <span class="math-container">$\alpha z \in Y$</span>, and as <span class="math-container">$Y$</span> is a convex cone, we have <span class="math-container">$2(\frac{1}{2} x + \frac{1}{2}\alpha z) = x + \alpha z \in Y$</span>. As <span class="math-container">$\alpha \ge 0$</span> and <span class="math-container">$x$</span> were arbitrarily, this shows that <span class="math-container">$z \in 0^+ Y$</span>.</p>
</blockquote>
<p>We then have the following Corollary whose proof follows from the previous two theorems.</p>
<blockquote>
<p><strong>Corr:</strong> Let <span class="math-container">$Y_1,\ldots, Y_n$</span> be a set of closed, convex and CRS production possibility sets. If <span class="math-container">$Y_1, \ldots, Y_n$</span> are positively semi-independent, then <span class="math-container">$Y = \sum_i Y_i$</span> is closed.</p>
</blockquote>
<p>Towards the interpretation of this condition. Assume for a moment that we only have two closed and convex CRS production possibility sets <span class="math-container">$Y_1, Y_2$</span> and assume that they are not positively semi-independent. This means that we can find <span class="math-container">$x \in Y_1$</span> and <span class="math-container">$y \in Y_2$</span> such that <span class="math-container">$x \ne 0$</span> and
<span class="math-container">$$
x + y = 0.
$$</span>
This means that <span class="math-container">$x \in Y_1$</span> and <span class="math-container">$y = -x = Y_2$</span>. In other words, we can use the first technology to produce  <span class="math-container">$x$</span> and then, in turn, the second technology to produce <span class="math-container">$-x$</span>. Intuitively, this corresponds to a reversible technology. For example, you use, labour, wood and nails to make a table (<span class="math-container">$x$</span>) and then, you can use this table as an inputs to make back the original inputs (labour, wood and nails) (<span class="math-container">$-x$</span>). In physics, this is called a <em>perpetual motion machine</em> (which <a href="https://en.wikipedia.org/wiki/Perpetual_motion" rel="nofollow noreferrer">violates the first or second law of theromodynamics</a>). 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.</p>

CC BY-SA 4.0
2021-06-03T15:00:12.323Z
Thanks for this thorough answer!
John Sturm
CC BY-SA 4.0
2021-06-03T08:20:47.093Z
2021-06-03T08:20:47.093Z
tdm
<p>Consider an economy with finitely many goods <span class="math-container">$I$</span> and factors <span class="math-container">$F$</span>. For each good <span class="math-container">$i$</span>, let <span class="math-container">$Y_i \subset \mathbb R^I \times \mathbb R^F_{\geq 0}$</span> denote a production set for <span class="math-container">$i$</span>. Assume each <span class="math-container">$Y_i$</span> has constant returns to scale and for all <span class="math-container">$(y,l)\in Y_i$</span>, <span class="math-container">$y_{j \neq i} \leq 0$</span>. Finally, denote by <span class="math-container">$Y \equiv \{ (\sum_{i \in I}y^i,\sum_{i \in I}l^i) \ | \ \forall i \in I, (y^i,l^i) \in Y_i\}$</span> the canonical aggregate production set.</p>
<p>It is common in GE theory to assume that <span class="math-container">$Y$</span> is closed.</p>
<p>Is it known what conditions on the sets <span class="math-container">$Y_i$</span> guarantee this?</p>
<p>To see that it is not sufficient to simply assume each <span class="math-container">$Y_i$</span> is closed, consider the following counterexample: There are two goods and one factor. Given one unit of labor, each firm can produce <span class="math-container">$n + \frac{n}{n+1}$</span> units of output per <span class="math-container">$n$</span> inputs of the other good used. By taking a limit as <span class="math-container">$n\to \infty$</span> for each firm, one may see that the closure of <span class="math-container">$Y$</span> contains <span class="math-container">$(y,l) = ((1,1),2)$</span></p>

CC BY-SA 4.0
2021-06-03T04:36:59.280Z
Your counterexample seems to violate constant returns to scale. So do we assume that or not?
Giskard
CC BY-SA 4.0
2021-06-03T08:47:38.143Z
An example that satisfies CRS can be found on page 4 [here](http://ftp.utia.cas.cz/pub/staff/studeny/concon-pdf.pdf), though it is not written in economic language.
Michael Greinecker
CC BY-SA 4.0
2022-01-28T18:52:48.430Z
For any who are interested, I found a reference that contains the main ideas of Michael's response. See Section 5 of Otani (1973), "Neo-Classical Technology Sets and Properties of Production Possibility Sets."
John Sturm
<general-equilibrium><aggregate-production><input-output><aggregate-supply>
Aggregation of the closure property of a production set
John Sturm
# Aggregation of the closure property of a production set

(Asked by: username_0 on 2021-06-03)

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 $(y,l) = ((1,1),2)$

**Tags:** general-equilibrium, aggregate-production, input-output, aggregate-supply
> **Comments:**
> **username_1**: Your counterexample seems to violate constant returns to scale. So do we assume that or not?  
> **username_2**: An example that satisfies CRS can be found on page 4 [here](http://ftp.utia.cas.cz/pub/staff/studeny/concon-pdf.pdf), though it is not written in economic language.  
> **username_0**: For any who are interested, I found a reference that contains the main ideas of Michael's response. See Section 5 of Otani (1973), "Neo-Classical Technology Sets and Properties of Production Possibility Sets."  

---

## 2 answers

### Answer by username_2 on 2021-06-03. Score: 7

Doing this more abstractly, let $Y\_j\subseteq\mathbb{R}^n$ be a production set for $j=1,\ldots,J$ and let
$$Y=Y\_1+Y\_2+\cdots+Y\_J=\{y\_1+y\_2+\cdots+y\_J|y\_j\in Y\_j, j=1,\ldots,J\}$$
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 $j=1,\ldots,J$ 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 $(y^1\_m,\ldots,y^J\_m)\in Y\_1\times\ldots\times Y\_J$ be such that $y\_m=\sum\_{j=1}^J y^j\_m$. 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 $(x^1\_m,\ldots,x^J\_m)$ by $x^j\_m=y^j\_m/\|(y^1\_m,\ldots,y^J\_m)\|$. The sequence has a converging subsequence $(z^1\_m,\ldots,z^J\_m)\to(z^1,\ldots,z^J)$ because it lies in the compact unit-sphere and $(z\_1,\ldots,z^J)\in Y\_1,\ldots, Y\_J$ because each $Y\_j$ is convex and contains $0$. Also, $\|(z^1,\ldots,z^J)\|=1$, so at least one coordinate must be nonzero. W.l.o.g. let $z^1\neq 0$. We have $$\sum\_{j=1}^J z^j=\lim\_{m\to\infty} \frac{y\_m}{\|(y\_m^1,\ldots,y\_m^J)|\|}=0$$
because $\|(y^1\_m,\ldots,y^J\_m)\|$ is unbounded and the convergent sequence $(y\_m)$ is bounded.
So $z^1-\sum\_{j=2}^J z^j=0$, but $z\_1\in Y$ and $\sum\_{j=2}^J z^j\in Y$ because each $Y\_j$ contains $0$, so this contradicts $Y\cap-Y=\{0\}$.

> **Comments:**
> **username_3**: 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 $Y_1\times \ldots \times Y_n$?  
> **username_2**: Yes, I mean the latter sequence in the product space.  
> **username_0**: 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.  
> **username_2**: @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.  

---

### Answer by username_3 on 2021-06-03. Score: 4

For a fully overview on the conditions for the sum of closed sets to be closed, see this note of [Kim Border](http://www.its.caltech.edu/%7Ekcborder/Notes/AsymptoticCones.pdf)

### 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](http://www.its.caltech.edu/%7Ekcborder/Notes/AsymptoticCones.pdf)) 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](https://en.wikipedia.org/wiki/Perpetual_motion)). 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.

> **Comments:**
> **username_0**: Thanks for this thorough answer!  

---

