Proving local non-satiation in arbitrary metric space

Question

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.

Answer 1

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.

Answer 2

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 $x'\succ x$.