<p>First, you need a vector space in order for convex combinations to be well-defined. However, not every metric on a vector space works. Indeed, under the <a href="https://en.wikipedia.org/wiki/Discrete_space" rel="nofollow noreferrer">discrete metric</a>, the result will trivially fail unless space consists of the point <span class="math-container">$0$</span> alone. What works is a metric on the vector space such that the vector operations of addition and scalar multiplication are continuous. Actually, no metric is needed and the result holds any <a href="https://en.wikipedia.org/wiki/Topological_vector_space" rel="nofollow noreferrer">topological vector space</a>. However, I doubt this is the level of generality the problem was posed at.</p>
<p>Anyways, here is the argument: Let <span class="math-container">$x$</span> be any point in <span class="math-container">$X_i$</span> that is not a global satiation point (there is not necessarily a unique satiation point because there might be no satiation point at all). Let <span class="math-container">$x^*$</span> be any point such that <span class="math-container">$x^*\succ_i x$</span>. Let <span class="math-container">$x_n=(n-1)/n\, x+1/n\, x^*$</span>. By strict convexity, <span class="math-container">$x_n\succ_i x$</span> for <span class="math-container">$n\ge 1$</span>. By the continuity of the vector operations, <span class="math-container">$\lim_{n\to\infty} x_n=x$</span>. Therefore, every neighborhood or ball around <span class="math-container">$x$</span> will contain a point of the form <span class="math-container">$x_n,$</span> which is strictly better. Therefore, <span class="math-container">$\succeq_i$</span> is locally non-satiated.</p>

CC BY-SA 4.0
2021-06-16T07:29:43.317Z
Thank you. But how does your conclusion that ‘every neighbourhood or ball around x will contain a point of the form x_n, which is strictly better (than x)’, follow from the fact that x_n converges to x?
Charles
CC BY-SA 4.0
2021-06-16T07:37:00.883Z
That is, how do we know that the distance between x_n and x is less than some epsilon>0?
Charles
CC BY-SA 4.0
2021-06-16T08:29:07.257Z
@Charles: Because this is part of the definition of $x_n$ converging to $x$.
VARulle
CC BY-SA 4.0
2021-06-16T09:33:40.940Z
@VARulle Thank you. I have attempted an answer below for the case in which we assume X is the positive real quadrant
Charles
CC BY-SA 4.0
2021-06-17T09:34:13.050Z
@Charles, this is not wrong, but unnecessarily complicated. By construction of $x_n$ the distance $||x_n-x||$ is $(1/n)\,||x-x^*||$, so it should be clear that for any $\epsilon>0$ we can find an $n$ with $||x_n-x||<\epsilon$. Any $n>||x-x^*||/\epsilon$ will do the job.
VARulle
CC BY-SA 4.0
2021-06-15T23:11:58.767Z
2021-06-16T08:25:17.427Z
2021-06-16T08:25:17.427Z
Michael Greinecker
<p>I have an answer for if we assume <span class="math-container">$X=\mathbb{R}^{2}_{+}$</span>.</p>
<p>Consider some <span class="math-container">$x:=(x_1, x_2) \in X$</span>, distinct from the satiation point (<span class="math-container">$x^{*}:=(x_1^{*}, x_2^{*})$</span>). Consider some <span class="math-container">$\epsilon&gt;0$</span>. By strict convexity, <span class="math-container">$x':=\bigg(\phi x_1+(1-\phi)x_1^{*}, \phi x_2+(1-\phi)x_2^{*}\bigg)\succ (x_1, x_2)  $</span> for all <span class="math-container">$\phi\in (0,1).$</span> Next, note that <span class="math-container">$||x'-x||$</span> is given by <span class="math-container">$\sqrt{(x'_1-x_1)^2+(x_2'-x_2)^2}=\sqrt{(\phi x_1+(1-\phi)x_1^{*}-x_1)^2+(\phi x_2+(1-\phi)x_2^{*}-x_2)^2}$</span>. Choose <span class="math-container">$\phi$</span> such that <span class="math-container">$(\phi x_1+(1-\phi)x_1^{*}-x_1)\leq \frac{\epsilon}{2}$</span> and <span class="math-container">$(\phi x_2+(1-\phi)x_2^{*}-x_2)\leq \frac{\epsilon}{2}$</span> (WLOG, we assume that <span class="math-container">$x_1^*&gt;x_1$</span> and <span class="math-container">$x_2^*&gt;x_2$</span>, so that <span class="math-container">$(\phi x_1+(1-\phi)x_1^{*}-x_1)$</span> and <span class="math-container">$(\phi x_2+(1-\phi)x_2^{*}-x_2)$</span> are strictly positive). This requires that <span class="math-container">$\phi \leq \frac{\frac{\epsilon}{2}+x_1-x_1^{*}}{x_1-x_1^{*}}$</span> and <span class="math-container">$\phi \leq \frac{\frac{\epsilon}{2}+x_2-x_2^{*}}{x_2-x_2^{*}}$</span>, equivalent, respectively, to  <span class="math-container">$1-\frac{\epsilon}{2(x_1^*-x_1)}$</span> and <span class="math-container">$1-\frac{\epsilon}{2(x_2^*-x_2)}$</span>. Note then that <span class="math-container">$||x'-x||\leq \frac{\epsilon}{\sqrt{2}}&lt;\epsilon.$</span></p>
<p>Thus, given any <span class="math-container">$x\in X$</span> and <span class="math-container">$\epsilon&gt;0$</span>, we can find some <span class="math-container">$x'$</span>, as defined above, with <span class="math-container">$\phi\in\bigg(0, \min\bigg\{1, 1-\frac{\epsilon}{2(x_1^*-x_1)},1-\frac{\epsilon}{2(x_2^*-x_2)} \bigg\}\bigg)—$</span>so that <span class="math-container">$x'\in B_\epsilon(x)—$</span>and such that <span class="math-container">$x'\succ x$</span>.</p>

CC BY-SA 4.0
2021-06-16T08:18:26.223Z
2021-06-16T10:32:57.900Z
2021-06-16T10:32:57.900Z
Charles
<p>I have a pure exchange economy where every consumption set <span class="math-container">$X_i$</span> is non-empty and convex and every preference relation <span class="math-container">$\succeq_{i}$</span> is strictly convex. I am asked to show that preferences are locally non-satiated at any consumption bundle different from the single ‘satiation point’ (i.e. the bundle which is strictly preferred to all other bundles in the consumption set; I have already proved that the satiation point, <span class="math-container">$x^*$</span>, is unique).</p>
<p>So far I have:</p>
<p>Consider an arbitrary <span class="math-container">$x_{i}\in X_{i}$</span>. Consider <span class="math-container">$\hat{x}:=\phi x_i+(1-\phi)x^*$</span>, <span class="math-container">$\phi\in(0,1)$</span>. Consider an arbitrary <span class="math-container">$\epsilon&gt;0$</span>. Now, choose <span class="math-container">$\phi$</span> such that the distance between <span class="math-container">$\hat{x}$</span> and <span class="math-container">$x_{i}$</span> is less than <span class="math-container">$\epsilon$</span>. That is, <span class="math-container">$\phi\in(0,1)$</span> such that <span class="math-container">$\phi x_i+(1-\phi)x^*&lt;\epsilon$</span>. I get that <span class="math-container">$\phi&lt;\frac{\epsilon+x_{i}-x^{*}}{x_{i}-x^{*}}$</span>.  We must have that <span class="math-container">$\phi\in(0, \min\{1, \frac{\epsilon+x_{i}-x^{*}}{x_{i}-x^{*}}\})$</span>. Also, note that, by strict convexity, <span class="math-container">$\hat{x}\succ x_{i}$</span>.</p>
<p>Beyond this point I am stuck. Any guidance would be much appreciated. Thank you.</p>

CC BY-SA 4.0
2021-06-15T20:37:48.603Z
How do you define strict convexity? And where do the arbitrary metric spaces come in?
Michael Greinecker
CC BY-SA 4.0
2021-06-15T20:42:43.933Z
Strict convexity: if two (distinct) bundles are weakly preferred to some third bundle, then any linear combination of those two bundles, with weights in $(0,1)$, is strictly preferred to the third.
Charles
CC BY-SA 4.0
2021-06-15T20:43:33.960Z
Arbitrary metric spaces come in because no assumption is being made as to the type of metric that $X$ is. i.e. we have not said that $X =\mathbb{R}^{N}_{+}$
Charles
CC BY-SA 4.0
2021-06-15T22:57:27.653Z
Convex combinations do not make sense in general metric spaces. What you need is a metric on a vector space such that the vector operations are continuous. The expression $\phi x_i+(1-\phi)x^*<\epsilon$ does not really make sense.
Michael Greinecker
<microeconomics><consumer-theory><general-equilibrium>
Proving local non-satiation in arbitrary metric space
Charles
# Proving local non-satiation in arbitrary metric space

(Asked by: username_0 on 2021-06-15)

I have a pure exchange economy where every consumption set $X\_i$ is non-empty and convex and every preference relation $\succeq\_{i}$ is strictly convex. I am asked to show that preferences are locally non-satiated at any consumption bundle different from the single ‘satiation point’ (i.e. the bundle which is strictly preferred to all other bundles in the consumption set; I have already proved that the satiation point, $x^\*$, is unique).

So far I have:

Consider an arbitrary $x\_{i}\in X\_{i}$. Consider $\hat{x}:=\phi x\_i+(1-\phi)x^\*$, $\phi\in(0,1)$. Consider an arbitrary $\epsilon>0$. Now, choose $\phi$ such that the distance between $\hat{x}$ and $x\_{i}$ is less than $\epsilon$. That is, $\phi\in(0,1)$ such that $\phi x\_i+(1-\phi)x^\*<\epsilon$. I get that $\phi<\frac{\epsilon+x\_{i}-x^{\*}}{x\_{i}-x^{\*}}$. We must have that $\phi\in(0, \min\{1, \frac{\epsilon+x\_{i}-x^{\*}}{x\_{i}-x^{\*}}\})$. Also, note that, by strict convexity, $\hat{x}\succ x\_{i}$.

Beyond this point I am stuck. Any guidance would be much appreciated. Thank you.

**Tags:** microeconomics, consumer-theory, general-equilibrium

> **Comments:**
> **username_1**: How do you define strict convexity? And where do the arbitrary metric spaces come in?  
> **username_0**: Strict convexity: if two (distinct) bundles are weakly preferred to some third bundle, then any linear combination of those two bundles, with weights in $(0,1)$, is strictly preferred to the third.  
> **username_0**: Arbitrary metric spaces come in because no assumption is being made as to the type of metric that $X$ is. i.e. we have not said that $X =\mathbb{R}^{N}_{+}$  
> **username_1**: Convex combinations do not make sense in general metric spaces. What you need is a metric on a vector space such that the vector operations are continuous. The expression $\phi x_i+(1-\phi)x^*<\epsilon$ does not really make sense.  

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## 2 answers

### Answer by username_1 on 2021-06-15. Score: 6

First, you need a vector space in order for convex combinations to be well-defined. However, not every metric on a vector space works. Indeed, under the [discrete metric](https://en.wikipedia.org/wiki/Discrete_space), the result will trivially fail unless space consists of the point $0$ alone. What works is a metric on the vector space such that the vector operations of addition and scalar multiplication are continuous. Actually, no metric is needed and the result holds any [topological vector space](https://en.wikipedia.org/wiki/Topological_vector_space). However, I doubt this is the level of generality the problem was posed at.

Anyways, here is the argument: Let $x$ be any point in $X\_i$ that is not a global satiation point (there is not necessarily a unique satiation point because there might be no satiation point at all). Let $x^\*$ be any point such that $x^\*\succ\_i x$. Let $x\_n=(n-1)/n\, x+1/n\, x^\*$. By strict convexity, $x\_n\succ\_i x$ for $n\ge 1$. By the continuity of the vector operations, $\lim\_{n\to\infty} x\_n=x$. Therefore, every neighborhood or ball around $x$ will contain a point of the form $x\_n,$ which is strictly better. Therefore, $\succeq\_i$ is locally non-satiated.

> **Comments:**
> **username_0**: Thank you. But how does your conclusion that ‘every neighbourhood or ball around x will contain a point of the form x_n, which is strictly better (than x)’, follow from the fact that x_n converges to x?  
> **username_0**: That is, how do we know that the distance between x_n and x is less than some epsilon>0?  
> **username_2**: @username_0: Because this is part of the definition of $x_n$ converging to $x$.  
> **username_0**: @username_2 Thank you. I have attempted an answer below for the case in which we assume X is the positive real quadrant  
> **username_2**: @username_0, this is not wrong, but unnecessarily complicated. By construction of $x_n$ the distance $||x_n-x||$ is $(1/n)\,||x-x^*||$, so it should be clear that for any $\epsilon>0$ we can find an $n$ with $||x_n-x||<\epsilon$. Any $n>||x-x^*||/\epsilon$ will do the job.  

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### Answer by username_0 on 2021-06-16. Score: 1

I have an answer for if we assume $X=\mathbb{R}^{2}\_{+}$.

Consider some $x:=(x\_1, x\_2) \in X$, distinct from the satiation point ($x^{\*}:=(x\_1^{\*}, x\_2^{\*})$). Consider some $\epsilon>0$. By strict convexity, $x':=\bigg(\phi x\_1+(1-\phi)x\_1^{\*}, \phi x\_2+(1-\phi)x\_2^{\*}\bigg)\succ (x\_1, x\_2) $ for all $\phi\in (0,1).$ Next, note that $||x'-x||$ is given by $\sqrt{(x'\_1-x\_1)^2+(x\_2'-x\_2)^2}=\sqrt{(\phi x\_1+(1-\phi)x\_1^{\*}-x\_1)^2+(\phi x\_2+(1-\phi)x\_2^{\*}-x\_2)^2}$. Choose $\phi$ such that $(\phi x\_1+(1-\phi)x\_1^{\*}-x\_1)\leq \frac{\epsilon}{2}$ and $(\phi x\_2+(1-\phi)x\_2^{\*}-x\_2)\leq \frac{\epsilon}{2}$ (WLOG, we assume that $x\_1^\*>x\_1$ and $x\_2^\*>x\_2$, so that $(\phi x\_1+(1-\phi)x\_1^{\*}-x\_1)$ and $(\phi x\_2+(1-\phi)x\_2^{\*}-x\_2)$ are strictly positive). This requires that $\phi \leq \frac{\frac{\epsilon}{2}+x\_1-x\_1^{\*}}{x\_1-x\_1^{\*}}$ and $\phi \leq \frac{\frac{\epsilon}{2}+x\_2-x\_2^{\*}}{x\_2-x\_2^{\*}}$, equivalent, respectively, to $1-\frac{\epsilon}{2(x\_1^\*-x\_1)}$ and $1-\frac{\epsilon}{2(x\_2^\*-x\_2)}$. Note then that $||x'-x||\leq \frac{\epsilon}{\sqrt{2}}<\epsilon.$

Thus, given any $x\in X$ and $\epsilon>0$, we can find some $x'$, as defined above, with $\phi\in\bigg(0, \min\bigg\{1, 1-\frac{\epsilon}{2(x\_1^\*-x\_1)},1-\frac{\epsilon}{2(x\_2^\*-x\_2)} \bigg\}\bigg)—$so that $x'\in B\_\epsilon(x)—$and such that $x'\succ x$.


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