# Does WARP still imply Compensated Law of Demand even if one of the goods is indivisible?

Consider 2 goods, $$x_1$$ and $$x_2$$. Let's assume $$x_1$$ is indivisible (we can only consume integer units of $$x_1$$). And suppose the consumer satisfies budget balance and homogeneity of degree 0, and WARP. Can we still say compensated law of demand holds?

According to proposition 2.F.1 in Mas-Colell, Whinston and Green (MWG):

If the demand function is

(a) homogeneous of degree zero and

(b) satisfies Walras' law (all wealth is spent),

then the demand function satisfies WARP if and only if it satisfies the compensated law of demand.

In MWG, the proof that (a), (b) and WARP imply the compensated law of demand uses only (b) and WARP. (It is only the converse - i.e. that (a), (b) and the compensated law of demand imply WARP - that requires condition (a).)

Thus, if your extra assumption about demand does not prevent the demand function from satisfying (b) and WARP, then the compensated law of demand holds by the proof in MWG.

Condition (b) can hold for demand functions where $$x_1$$ is integer-valued as can WARP.

If you had assumed both goods are only consumed in integer values, then condition (b) would be violated because it would not be possible to spend all wealth for all combinations of prices and wealth. Then, one would have to see if the proof of the relevant part of proposition 2.F.1 in MWG can be amended. It turns out that it can, if instead of Walras' Law we assume just that $$p\cdot x(p,w)\leq w$$ for any prices $$p$$ and wealth $$w$$. Call this condition (b').

The proof boils down to showing that for any compensated price change from $$(p,w)$$ to $$(p',w')=(p',p'\cdot x(p,w))$$ we have, when $$x(p',w')\neq x(p,w)$$, that:

$$p'\cdot[x(p',w')-x(p,w)]\leq 0 \tag{2.F.3}$$

and

$$p\cdot[x(p',w')-x(p,w)]>0\tag{2.F.4}$$

Consider $$(2.F.3)$$. We have $$p'\cdot x(p',w')\leq w'$$ by condition (b') and $$p'\cdot x(p,w)=w'$$ since it is a compensated price change. Thus $$(2.F.3)$$ holds. Consider $$(2.F.3)$$. Because $$p'\cdot x(p,w)=w'$$, $$x(p,w)$$ is affordable at $$(p',w')$$. and so by WARP $$x(p',w')$$ must not be affordable at $$(p,w)$$. Thus $$p\cdot x(p',w')>w$$. Also $$p\cdot x(p,w)\leq w$$ by (b'). Thus $$(2.F.4)$$ holds.