What's the most straight-forward way to prove Walras's Law? [migrated]

Walras' Law states that summation of pi Ei(p) = 0 for all pi. We define Ei(p) = xi(p) - qi(p) - Ri. What are the next steps that I should take?

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migrated to math.stackexchange.com by Jason B♦May 1 '12 at 20:01

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I don't quite know what your notation is. The proof starts by asserting LNS preferences and claiming walras' law, $\forall p,w$ and , $x \in x(p,w), p\cdot x=w$ The proof is almost always handled by contradiction. You can see most any micro textbook for the full proof. A good start would be to define your assumptions (LNS?) and the various functions you've specified (you'd have to do that for a proper proof anyway.) –  Jason B Nov 14 '11 at 4:30
@Jason B - what's LNS? –  Patience Nov 15 '11 at 17:59
Local non-satiation. It's the claim that, for any point $x$ and any number $\epsilon>0$, there exists a $x'$ in the $\epsilon$-neighbourhood of $x$ such that $x'$ is strictly preferred to $x$. –  Zermelo Nov 15 '11 at 18:59
thanks @user68! –  Patience Nov 15 '11 at 21:56
@Patience the most straightforward proof of Walras' Law requires one to assume LNS preferences and little more (it is implicit in Zermelo's answer). –  Jason B Nov 16 '11 at 3:46