<p>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 $\alpha_0*e&lt;\alpha(x')*e&lt;\alpha_1*e$ (with $&lt;$ I mean strictly preferred). Remember that, by monotonicity, if $y&gt;&gt;x$, then $y$ is strictly preferred to $x$. Therefore, we only need to find some $\alpha_0$ and $\alpha_1$ such that $\alpha_0&lt;\alpha(x')&lt;\alpha_1$ for all $x'$. </p>

<p>This is because the set $\{x':||x'-x||\le\epsilon\}$ is bounded and $x'∼\alpha(x')*e$. </p>

<p>(Note that with local non-satiation the result does not immediately follow).</p>

<p>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. </p>

CC BY-SA 3.0
2016-06-22T02:22:43.823Z
Alvaro, this helps quite a bit. Thanks! Are you economics graduate student? :)
Frank Swanton
CC BY-SA 3.0
2016-06-22T02:25:49.717Z
One point of clarification: so where does alpha(x') is on? It says a compact subset of R+ but the Figure points to the diagonal ray and says compact subset of Z.
Frank Swanton
CC BY-SA 3.0
2016-06-22T02:37:32.143Z
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
CC BY-SA 3.0
2016-06-22T02:44:54.783Z
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
CC BY-SA 3.0
2016-06-22T04:37:49.537Z
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
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2016-06-22T04:46:38.630Z
Yes, you can also have an open subset $(\alpha_0,\alpha_1)$ that  contains alpha(x') for all x' (in fact, you can use the same $\alpha_0$ and $\alpha_1$). MWG define a closed one with those values because it is important for the proof (a convergent subsequence exists).
Belisario
CC BY-SA 3.0
2016-06-22T05:06:38.987Z
Your answer is fine, but see that monotonicity assures boundedness immediately. For any neighborhood, you can find $x_0$ and $x_1$ such $x_0 < x' < x_1$ (< means strictly preferred). Just draw a square around the circle-shaped neighborhood and take values from the north-east and south-west. We know there is a unique $\alpha_0$ and a unique $\alpha_1$ such that $x_1 ∼ \alpha_0*e$ and $x_0 ∼ \alpha_1*e$. Therefore, we have $ \alpha_0*e <  \alpha(x')*e <  \alpha_1*e$. By monotonicity, we know this implies $ \alpha_0*e <<  \alpha(x')*e <<  \alpha_1*e$ or $ \alpha(x') \in [\alpha_0,\alpha_1]$.
Belisario
CC BY-SA 3.0
2016-06-23T13:13:02.897Z
Alvaro, thanks for the comments. Let me take a closer look at make few comments!
Frank Swanton
CC BY-SA 3.0
2016-06-23T13:14:17.857Z
Are you starting this fall, Alvaro?
Frank Swanton
CC BY-SA 3.0
2016-06-21T22:54:29.093Z
2016-06-21T22:54:29.093Z
Belisario
<p>I am reading MWG's explanation in Chapter 3 when showing continuous preference relation implies the existence of continuous utility function.</p>

<p>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.</p>

<p>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.</p>

<p><strong><em>The part I am having trouble is when they use monotonicity to show this compact set. The exact excerpt is:</em></strong></p>

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

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

<p>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?</p>

<p>Please help. Thank you.</p>

<mathematical-economics><consumer-theory><preferences><choice-theory>
Continuous Preference Relation Imply Continuous Utility Fn Existence
Frank Swanton
# Continuous Preference Relation Imply Continuous Utility Fn Existence

(Asked by: username_0 on 2016-06-21)

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?

Please help. Thank you.

**Tags:** mathematical-economics, consumer-theory, preferences, choice-theory

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## 1 answers

### Answer by username_1 on 2016-06-21. Score: 2 | Accepted answer

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 $x'∼\alpha(x')\*e$.

(Note that with local non-satiation the result does not immediately follow).

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.

> **Comments:**
> **username_0**: Alvaro, this helps quite a bit. Thanks! Are you economics graduate student? :)  
> **username_0**: One point of clarification: so where does alpha(x') is on? It says a compact subset of R+ but the Figure points to the diagonal ray and says compact subset of Z.  
> **username_0**: 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.  
> **username_0**: My other curiosity and silliness is that why does alpha-0 and -1 have to be compact. Can't they be just open?  
> **username_1**: 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+.  
> **username_1**: Yes, you can also have an open subset $(\alpha_0,\alpha_1)$ that  contains alpha(x') for all x' (in fact, you can use the same $\alpha_0$ and $\alpha_1$). MWG define a closed one with those values because it is important for the proof (a convergent subsequence exists).  
> **username_1**: Your answer is fine, but see that monotonicity assures boundedness immediately. For any neighborhood, you can find $x_0$ and $x_1$ such $x_0 < x' < x_1$ (< means strictly preferred). Just draw a square around the circle-shaped neighborhood and take values from the north-east and south-west. We know there is a unique $\alpha_0$ and a unique $\alpha_1$ such that $x_1 ∼ \alpha_0*e$ and $x_0 ∼ \alpha_1*e$. Therefore, we have $ \alpha_0*e <  \alpha(x')*e <  \alpha_1*e$. By monotonicity, we know this implies $ \alpha_0*e <<  \alpha(x')*e <<  \alpha_1*e$ or $ \alpha(x') \in [\alpha_0,\alpha_1]$.  
> **username_0**: Alvaro, thanks for the comments. Let me take a closer look at make few comments!  
> **username_0**: Are you starting this fall, Alvaro?  

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