# If $x \succsim_i x_i^*$ then $p \cdot x_i\ge w_i.$ (MWG 16.C.2)

(MWG 16.C.2) Suppose that the preference relation $$\succsim_i$$ is locally nonsatiated and that $$x_i^*$$ is maximal for $$\succsim_i$$ in set $$\{x_i \in X_i: p \cdot x_i \le w_i\}$$. Prove that the following property holds: "If $$x \succsim_i x_i^*$$ then $$p \cdot x_i\ge w_i.$$"

Locally nonsatiation means that for every $$x_i \in X_i$$ and $$\epsilon >0$$, there exists $$x_i' \in X_i$$ such that $$||x_i - x_i'|| < \epsilon$$ and $$x_i' \succ x_i$$. I think that I need to prove that if $$x_i \sim_i x^*_i$$ for $$x_i \not= x_i^*$$, then it is possible that $$p\cdot x_i = w_i$$ because I already know that if $$x_i \succ x_i^*$$, then $$p\cdot x_i > w_i$$.

I am stuck and cannot proceed. Can anyone give me some hint for this question?

Hint: Prove by contradiction. That is, suppose $$x_i\succsim_ix_i^*$$ but $$p\cdot x_i, show that this would lead to a contradiction of $$x_i^*$$ being maximal for the feasible set.