$\succsim$ preorder on X being continuous imply lower contour set closed

Solution 1:

You appear in effect to be assuming that $\succsim$ is the same as ordinary $\ge$. This need not be the case. In fact, the whole point of the argument is that the lower contour sets for any continuous preorder on $X$ are closed, so that in this sense all continuous preorders on $X$ behave like the familiar natural order $\ge$. Of course the steps of the argument will seem reasonable if you look only at that one order, but in order actually to understand the argument, you have to see why it works for all continuous preorders on $X$. For instance, in (3) it need not be true that $L(x)$ ‘is exactly a collection of real numbers $y$ such that it is less or equal to some $x$ on the real line’. That’s true for $\ge$, but it’s not in general true for a continuous preorder $\succsim$, so you can’t use it to justify the conclusion. $L(x)$ is actually the set of all real numbers $y$ such that $y\in X$ and $x\succsim y$; if $X=[0,2]$, $x=1$, and $\succsim$ is $\le$, $L(x)=L(1)=[1,2]$ is in fact the set of elements of $X$ that are greater than or equal to $1$ on the real line.

This is essentially the same mistake that Henno Brandsma discussed in the third paragraph of his answer to this question. It’s still a mistake even if you’re only trying to explain the result to beginners, since it obscures the real point of the theorem.

(2) is just a direct application of the definition of continuity of a preference relation. Suppose that $x\succsim y^n$ for each $n\in\Bbb Z^+$, where $\lim_{n\to\infty}y^n=y$. For each $n\in\Bbb Z^+$ let $x^n=x$; then $\langle\langle x^n,y^n\rangle:n\in\Bbb Z^+\rangle$ is a sequence of pairs of elements of $X$ such that $x^n\succsim y^n$ for each $n\in\Bbb Z^+$. Clearly $\lim_{n\to\infty}x^n=x$, since the sequence is constant at $x$, so $x\succsim y$ by the definition of continuity of $\succsim$. For expository purposes you could give a concrete example, but it would be best not to use $\ge$, precisely to avoid giving the impression that it’s the only continuous preorder.

For (3) you want to prove that $L(x)$ is closed, so you should start with an arbitrary $y$ in the closure of $L(x)$ and try to show that $y\in L(x)$. If $y$ is in the closure of $L(x)$, there is a sequence $\langle y^n:n\in\Bbb Z^+\rangle$ in $L(x)$ such that $y=\lim_{n\to\infty}y^n$. Then for each $n\in\Bbb Z^+$ we have $x\succsim y^n$, since $y^n\in L(x)$, and it’s an immediate consequence of (2) that $x\succsim y$. By definition that means that $y\in L(x)$, and we’re done: every limit point of $L(x)$ is already in $L(x)$, so $L(x)$ is closed.