$\succsim$ is preorder (i.e. preference relation) on X that is continuous.
This implies the lower contour set is closed.
Would you please share your 2 cent on my parenthesis explanation (e.g. line of reasoning)? The actual proof requires only three lines, but I would like to see if my thinking is correct and as if I were explain this to someone in undergraduate analysis class. Thanks!
My proof:
(1) Recall Definition 3.C.1(Modified for $\mathbb{R}$):
The preference relation $\succsim$ on $X\subset\mathbb{R_+}$ is continuous if it is preserved under limits. That is, for any sequence of pairs $\{(x^n, y^n)\}^\infty_{n=1}$ with $x^n \succsim y^n$ for all $n$, $x = \lim_{n \rightarrow \infty} x^n$, and $y = \lim_{n \rightarrow \infty} y^n$, we have $x \succsim y$.
(In words, if you have a sequence of reals $x^n$ and $y^n$ where each term in $x^n$ is greater or equal to that of in $y^n$ and $x^n\rightarrow x$, $y^n\rightarrow y$, then you have $x\succsim y$. Notice here, to facilitate understanding, I basically equated the preference relation as the in\equality on reals, thus the expression "greater or equal". An example is $x=2,y=1$. Essentially, look at the equivalence class of 2 and 1 on real line with the restriction that it is not any sequence converging to these numbers but need to narrow down to the group of convergent sequences where each term in $x^n$ is greater or equal to that of in $y^n$.)
(2) The above definition implies the following: for any sequence of pairs $\{(x^n, y^n)\}^\infty_{n=1}$ with $x\succsim y^n$ $\forall n$, where $x^n=x$ $\forall n$ with $y^n$ converging to $y$, we have $x\succsim y$.
((1) certainly implies (2), because again think of the example, $x^n=2, 2, 2, \dots$ and $y^n$ being any member of the equivalence class of 1 with the restriction each term has to be less or equal to 2. How does the set of the sequences $y^n$ satisfying this condition actually look like on the real line? Each term has to be less or equal to 2 but we are in the nonnegative commodity space $\mathbb{R_+}$, so we have a set of sequences bounded in the interval $[0,2]$ but convergent to the number 1. This example is still valid for any pair of sequences of reals $x^n\geq y^n$ where $y^n$, the sequence of reals is a bounded sequence in $[0,x]$ converging to any a point $y\in[0,x]$ while $x^n=x,x,x,\dots$, we have $x\geq y$.)
(3) Define the lower contour set to be $L(x)=\{y\in X:x\succsim y\}$. Then, (2) implies $L(x)$ is closed.
(In (2), we attained that for any pair of sequences of reals $x^n\geq y^n$ where $y^n$, the sequence of reals is a bounded sequence in $[0,x]$ converging to any a point $y\in[0,x]$ while $x^n=x,x,x,\dots$, we have $x\geq y$. The lower contour set is exactly a collection of real numbers $y$ such that it is less or equal to some $x$ on the real line. Then for any real number $y\in[0,x]$, you have a sequence of reals bounded by $0$ and $x$ where it converges to that $y$.Then, $L(x)$ contains all its limit points. Hence, it is closed.)
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.