# On closedness: commodity space

In Intriligator (2002, p. 143) we find the following statement:

\begin{align} C = \{(x_1,x_2,\ldots,x_n) : x_j \geq 0,~j=1,2,\ldots,n\} \subset [0,\infty)^n \end{align} Thus commodity space is the nonnegative orthant of Euclidean $n$-space, a closed, convex set.

I'm rather confused why $C$ is supposed to be closed, because $x_j$ is not bounded from above. However, we may argue that $C$ is closed, because its complement \begin{align} C^c \subset (-\infty,0)^n \end{align} is open.

• Still, isn't it appropriate to say that $C$ is half-closed? Am I hairsplitting here?

A closed set does not need to be bounded. For instance, the set $[0,\infty)$ is closed but unbounded.
Formally, a set is closed if it contains all its limit points. You can easily verify that it is the case for your $C$. Take a sequence of elements $x^m=(x_1^m,\cdots,x_n^m) \in C$ that converges to a vector $x^{*}=(x_1^{*},\cdots,x_n^{*})$ under any appropriate topology. The $j$-th coordinate of $x^{*}$, $x_j^{*}$, is the limit of the sequence $x_j^{m}$. Since all $x_j^{m}$ are nonnegative real numbers, so is the case for the limit $x_j^{*}$. Since this is true for all $j$, $x^{*}$ also belongs to $C$.
• Ok, I'd like to provide a simple example to make sure, that I got the concept. Let $X=(0,1]$ and fix $x^k=1/k$. Now $X$ is open, because $\lim_{k\to\infty} x^k = 0 \not\in X$? However, $Y=[0,1]$ is closed? – clueless Dec 5 '15 at 14:58
• $X$ is not closed, you are right, because of the example you mentionned. But $X$ is not open either because it contains 1, which is one of its boundary points: the set $(0,1)$ is open, but not $(0,1]$. I suggest you skim through an introductory topology textbook to think of the definitions of openness and closedness again. – Oliv Dec 5 '15 at 15:11
• @clueless: A set is closed if it contains all its limit points. The set $X=(0,1]$ is not closed because one of its limit points, $0$, is not contained in the set $X$. However, $[0,\infty)$ is closed because "$\infty$" is not a limit point of the set (note that "$\infty$" is not a real number). – Herr K. Dec 6 '15 at 4:39