<p>A closed set does not need to be bounded. For instance, the set $[0,\infty)$ is closed but unbounded.</p>

<p>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 $C$.</p>

CC BY-SA 3.0
2015-12-05T14:58:51.787Z
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
CC BY-SA 3.0
2015-12-05T15:11:09.540Z
$X$ 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
CC BY-SA 3.0
2015-12-05T18:50:19.887Z
...That is to say, attaching infinity in the set of real makes the set "even more closed" because it is more than certain that all real numbers will be in there, since infinity exceeds them all (informally or in Abraham Robinson's rigorous "non-standard" approach). @clueless. Remember that we say "a continuous function has global extrema if the domain is closed _and_ bounded". If the properties went together, we would not have to invoke them both.
Alecos Papadopoulos
CC BY-SA 3.0
2015-12-06T04:39:09.290Z
@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.
CC BY-SA 3.0
2015-12-05T12:21:25.313Z
2015-12-05T18:58:26.260Z
2015-12-05T18:58:26.260Z
Oliv
<p>In <a href="http://bookstore.siam.org/cl39/" rel="nofollow">Intriligator (2002, p. 143)</a> we find the following statement:</p>

<blockquote>
  <p>\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.</p>
</blockquote>

<p>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.</p>

<ul>
<li>Still, isn't it appropriate to say that $C$ is <em>half</em>-closed? Am I hairsplitting here?   </li>
</ul>

<microeconomics><consumer-theory><theory>
On closedness: commodity space
clueless
# On closedness: commodity space

(Asked by: username_0 on 2015-12-05)

In [Intriligator (2002, p. 143)](http://bookstore.siam.org/cl39/) 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?

**Tags:** microeconomics, consumer-theory, theory

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## 1 answers

### Answer by username_1 on 2015-12-05. Score: 4

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 $C$.

> **Comments:**
> **username_0**: 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?  
> **username_1**: $X$ 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.  
> **username_2**: ...That is to say, attaching infinity in the set of real makes the set "even more closed" because it is more than certain that all real numbers will be in there, since infinity exceeds them all (informally or in Abraham Robinson's rigorous "non-standard" approach). @clueless. Remember that we say "a continuous function has global extrema if the domain is closed _and_ bounded". If the properties went together, we would not have to invoke them both.  
> **username_3**: @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).  

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