<p>It appears that you are conflating a "closed interval" with a "closed set". Certainly, (multidimensional) closed intervals are closed sets, but the concept of a closed set is wider. </p>

<p>A closed set is, well, only closed, and not necessarily bounded. </p>

<p>I understand that the everyday-meanings of the terms "closed" and "bounded" can indicate that "being closed is a stronger property than being bounded" (i.e. that being closed implies being bounded), but this is not true in mathematics. Neither property implies the other.  </p>

<p>Yes, a set can be bounded and yet it may not be closed. Standard example: Consider the set of <em>rational</em> numbers  in $[1,3]$ . This is a bounded set. Now consider the sequence of rational numbers</p>

<p>$$x_{n+1} = \frac {x_n}{2} +\frac {1}{x_n},\;\; x_0 = 3$$</p>

<p>The sequence remains in the $[1,3]$ interval, but its <em>limit</em> is $\sqrt 2$, which is not rational. So this set of rationals in $[1,3]$ is bounded but it is not closed, because it does not contain a limit point of a sequence that can be constructed in it, stay in it, except its limit point.</p>

<p>So, what "closed set" means then in mathematics? There are various ways to look at it:</p>

<p>A closed set is a set whose complement is open.<br>
A closed set is a set that coincides with its "closure".<br>
A closed set is a set that contains all its limit points.</p>

<p>Try them out mentally and stick with what is more intuitive to you.</p>

CC BY-SA 3.0
2016-03-14T19:12:20.237Z
"A closed interval is a closed and bounded set, i.e. it is a compact set". Being closed and bounded is necessary but not sufficient for a set to be a closed interval (I am sure that you know this and that it is not a mistake, but this formulation is a bit ambiguous and might be understood by some readers as a definition instead of a necessary condition).
Oliv
CC BY-SA 3.0
2016-03-14T22:18:48.120Z
@Oliv Thanks for spotting this. I deleted the sentence since it was not central to the issue.
Alecos Papadopoulos
CC BY-SA 3.0
2016-03-14T15:45:04.297Z
2016-03-14T22:18:07.727Z
2016-03-14T22:18:07.727Z
Alecos Papadopoulos
<p>Suppose consumption set $X=R_{+}^N$. According to the definition preferences are continuous if for any $x\in X$ sets $(y\in X:x\succeq y)$ and $(y\in X:y\succeq x)$ are closed. It is clear that preferences represented by utility function $u(k_{1},k_{2})=k_{1}+k_{2}$ are continuous. But now lets take some bundle x=(1,1). Then set $(y\in X:y\succeq (1,1))$ is not closed as it approaches infinity. My question is how to handle with this contradiction?</p>

CC BY-SA 3.0
2016-03-14T15:17:43.280Z
I think you may be confusing **boundedness** with **closedness**. Note in particular that the half-open interval $[1,\infty)$ *is closed*, although it is *not bounded*.
Herr K.
<mathematical-economics><preferences>
Continuous preferences definition
Bogdan
# Continuous preferences definition

(Asked by: username_0 on 2016-03-14)

Suppose consumption set $X=R\_{+}^N$. According to the definition preferences are continuous if for any $x\in X$ sets $(y\in X:x\succeq y)$ and $(y\in X:y\succeq x)$ are closed. It is clear that preferences represented by utility function $u(k\_{1},k\_{2})=k\_{1}+k\_{2}$ are continuous. But now lets take some bundle x=(1,1). Then set $(y\in X:y\succeq (1,1))$ is not closed as it approaches infinity. My question is how to handle with this contradiction?

**Tags:** mathematical-economics, preferences

> **Comments:**
> **username_1**: I think you may be confusing **boundedness** with **closedness**. Note in particular that the half-open interval $[1,\infty)$ *is closed*, although it is *not bounded*.  

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## 1 answers

### Answer by username_2 on 2016-03-14. Score: 3 | Accepted answer

It appears that you are conflating a "closed interval" with a "closed set". Certainly, (multidimensional) closed intervals are closed sets, but the concept of a closed set is wider.

A closed set is, well, only closed, and not necessarily bounded.

I understand that the everyday-meanings of the terms "closed" and "bounded" can indicate that "being closed is a stronger property than being bounded" (i.e. that being closed implies being bounded), but this is not true in mathematics. Neither property implies the other.

Yes, a set can be bounded and yet it may not be closed. Standard example: Consider the set of *rational* numbers in $[1,3]$ . This is a bounded set. Now consider the sequence of rational numbers

$$x\_{n+1} = \frac {x\_n}{2} +\frac {1}{x\_n},\;\; x\_0 = 3$$

The sequence remains in the $[1,3]$ interval, but its *limit* is $\sqrt 2$, which is not rational. So this set of rationals in $[1,3]$ is bounded but it is not closed, because it does not contain a limit point of a sequence that can be constructed in it, stay in it, except its limit point.

So, what "closed set" means then in mathematics? There are various ways to look at it:

A closed set is a set whose complement is open.  
A closed set is a set that coincides with its "closure".  
A closed set is a set that contains all its limit points.

Try them out mentally and stick with what is more intuitive to you.

> **Comments:**
> **username_3**: "A closed interval is a closed and bounded set, i.e. it is a compact set". Being closed and bounded is necessary but not sufficient for a set to be a closed interval (I am sure that you know this and that it is not a mistake, but this formulation is a bit ambiguous and might be understood by some readers as a definition instead of a necessary condition).  
> **username_2**: @username_3 Thanks for spotting this. I deleted the sentence since it was not central to the issue.  

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