# Local nonsatiation

Suppose that $$x^*$$ satisfies $$x^*\succsim x$$ for $$\forall x\in\{{x∈X|p·x\leq m}\}$$.

How can we prove that $$x\succsim x^*$$ $$\Rightarrow$$ $$p·x≥m$$ if $$\succsim$$ is locally nonsatiated?

My idea for this goes like this but I am not sure:

Suppose that there exists $$x' ∈ X$$ such that $$x'\succsim x^∗$$ and $$p · x' < m.$$ Since we have local nonsatiation, for any $$\epsilon>0$$ there exists $$x''\in X$$ such that $$||x''−x'||≤\epsilon$$ and $$x''\succ x'$$. By transitivity of $$\succsim$$ and by $$x''\succ x'\succsim x^*$$. The former implies, by $$p·x' < m$$, that we can choose $$\epsilon$$ small enough such that $$p·x'' < m$$ and hence $$x''\in\{{x∈X|p·x\leq m}\}$$. This implies, by the definition of $$x^∗$$, that $$x^∗\succsim x''$$, a contradiction.

• Your idea is largely correct, but there are a few typos. (1) In the 2nd sentence of the last paragraph, $x$ should have been $x'$; (2) where transitivity is invoked, the chain should have been $x''\succ x'\succsim x^*$; and (3) there's an extraneous "$\forall x \in$" in the second last sentence specifying the set to which $x''$ belongs. Mar 1 '21 at 19:31
• I fixed them. I hope that it is correct now. Thank you for your input! Mar 1 '21 at 19:42

By local nonsatiation there exists a sequence $$x_n\rightarrow x$$ with $$x_n\succ x\,\,\forall n$$. By transitivity $$x_n\succ x^*\,\,\forall n$$. By contraposition $$p\cdot x_n>m\,\,\forall n$$. By continuity of the dot product $$p\cdot x\ge m$$.