Assignment:
Eilon contacts Michael, claiming that the second direction in the proof of Theorem (if the system of equations given by Equations is tight for x, then x is the nucleolus) does not hold: “Denote the nucleolus by x*, and let x be an imputation,” explains Eilon. “Denote y = x* − x. Then y(N) = 0. Let α1 be the greatest value of the excesses at θ(x*). Then for every coalition S,
y(S) = (x* − x)(S) = e(S, x) − e(S, x*).
Since x* is the nucleolus, and since α1 is the greatest value of the excesses, it must be the case that e(S, x) ≥ 0 for every coalition S ∈ D(α1, x*). In other words, y(S) ≥ 0 for every such coalition. Since in the first part of the theorem, it is proved that relative to the nucleolus, the system of equations is tight, one deduces that y(S) = 0 for every S ∈ D(α1, x*). Continuing by induction to the next level, one concludes that y(S) = 0 for every coalition S, which implies that x = x*. But this cannot hold since x was chosen arbitrarily.” Is Eilon correct? If not, where is the flaw in his argument?
Provide complete and step by step solution for the question and show calculations and use formulas.