MWG Proposition 2.F.1 Is homogeneity of degree zero necessary in proposition 2.F.1?
It seems like the proof does not assume homogeneity of degree zero to establish the proposition.
If this is true, it seems that homogeneity is not required to establish that the Slutsky matrix is negative semidefinite (only required assumptions are differentiability and Walras' law?)

Thank you so much for your help!
In between eq. $(2.F.2)$ and $(2.F.3)$ we read


  ...Walras's law tells us that $w' = p'\cdot x(p',w')$


Assume now that homogeneity of degree zero does not hold. Then we have, $a&gt;0$

$$x(ap,aw) \neq x(p,w)$$

see definition $2.E.1$

Then we can set $p'=ap,w'=aw$ to examine the case $x(p',w') \neq x(p,w)$.

But then Walras' law would imply

$$w' = p'\cdot x(p',w') \implies aw = ap\cdot x(ap,aw) \implies w = p\cdot x(ap,aw)$$

while also, again by Walras' law we have

$$w = p\cdot x(p,w)$$

So under the assumption that Walras' law holds but that we do not have homogeneity of degree zero we end up saying

$$p\cdot x(ap,aw) = p\cdot x(p,w), \,\,\, x(ap,aw) \neq x(p,w)$$

I believe you can contemplate the situation further from here.