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 $C$.