Continuous preferences definition

Question

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?

Answer 1

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.