# Proving local non-satiation in arbitrary metric space

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.

• How do you define strict convexity? And where do the arbitrary metric spaces come in? Jun 15 '21 at 20:37
• 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. Jun 15 '21 at 20:42
• 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 $X =\mathbb{R}^{N}_{+}$ Jun 15 '21 at 20:43
• 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. Jun 15 '21 at 22:57

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, 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. 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.

• 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? Jun 16 '21 at 7:29
• That is, how do we know that the distance between x_n and x is less than some epsilon>0? Jun 16 '21 at 7:37
• @Charles: Because this is part of the definition of $x_n$ converging to $x$. Jun 16 '21 at 8:29
• @VARulle Thank you. I have attempted an answer below for the case in which we assume X is the positive real quadrant Jun 16 '21 at 9:33
• @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. Jun 17 '21 at 9:34

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 $$x'\succ x$$.