# Continuous Preference Relation Imply Continuous Utility Fn Existence

I am reading MWG's explanation in Chapter 3 when showing continuous preference relation implies the existence of continuous utility function.

First, the authors show $u(.)$ is continuous by using the definition that the image under $u(.)$ of a convergent sequence is convergent. Consider a sequences $x_n\rightarrow x$. They first claim that $u(x_n)$ must have a convergent subsequence.

I understand the big picture: Since $x_n$ converges to x, for some large N, $u(x_n)$ must all lie in some compact set, and any infinite sequence in a compact set must have a convergent subsequence.

The part I am having trouble is when they use monotonicity to show this compact set. The exact excerpt is:

"By monotonicity, for any $\epsilon>0$, $\alpha(x')$ lies in a compact subset of $\mathbb{R_+}$, [$\alpha_0,\alpha_1$], for all $x'$ such that $\parallel x'-x\parallel$$\leq0$."

Here I used $u(.)$ and $\alpha(.)$ interchangeably to represent the utility function. Can somebody elaborate the above claim just little more in detail please?

I understand monotonicity implies local nonsatiation, hence, in any given small ball, you always have some bundle that you prefer that to x. Part of my confusion comes from the Figure they present, which is they put the bundle x on the indifference curve ($\{y\in X:y\sim x$}). But isn't $\alpha(x')$ on the diagnoal line Z?

I think you're overthinking this proof. MWG only say that such a compact set exists because, by monotonicity, you can always find $\alpha_0$ and $\alpha_1$ such that $\alpha_0*e<\alpha(x')*e<\alpha_1*e$ (with $<$ I mean strictly preferred). Remember that, by monotonicity, if $y>>x$, then $y$ is strictly preferred to $x$. Therefore, we only need to find some $\alpha_0$ and $\alpha_1$ such that $\alpha_0<\alpha(x')<\alpha_1$ for all $x'$.
This is because the set $\{x':||x'-x||\le\epsilon\}$ is bounded and $x'∼\alpha(x')*e$.
You're right, $\alpha(x')*e$ is on the diagonal line Z and $x'$ is not (note that $\alpha(x')$ is mutiplied by $e$, a vector filled by 1's). However, working with $\alpha(x')$ reduces the complexity of the problem. Instead of working with two (or more) dimensions, we just work with one (the diagonal line Z or one axis). We can do that because for any element $x\in \Bbb R_+$, there is $x∼\alpha(x)*e$ which is in the diagonal line.
• Acutally i think i figured. Please correct me if I am wrong. Basically think of Z as real line if you treat the commodity space as $\mathbb{R^+}&. Then basically it is saying, any x' that lies in a small neighborhood of x, it should have a corresponding alpha(x') and because the ordering property and completeness of R, you can always find this close and bounded interval. – Frank Swanton Jun 22 '16 at 2:37 • My other curiosity and silliness is that why does alpha-0 and -1 have to be compact. Can't they be just open? – Frank Swanton Jun 22 '16 at 2:44 • alpha(x') is on R+. Actually, you can forget the diagonal line Z and focus on the horizontal axis. MWG used the line Z because$\alpha(x')*e\$ means that the vector has a value of alpha(x') in the horizontal axis and also a value of alpha(x') in the vertical axis. Therefore, whatever you do in the line Z is valid for R+. – Belisario Jun 22 '16 at 4:37