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Proof of Proposition 2.4

Proof.
If E ⊆ F, then as we observed earlier, K_i(E) ⊆ K_i(F), so

K1N(E) =

∩ i ∈ N

Ki(E) =

∩ i ∈ N

Ki(F) = K1N(F)

If we now set E′ = Kⁿ_N(E) and F′ = Kⁿ_N(F), then by the argument just given we have

Kn+1N(E) = K1N(E′) ⊆ K1N(F′) = Kn+1N(F)

so we have m^th level mutual knowledge for every n ≥ 1.

Hence if ω ∈

∞ ∩ n=1

KnN(E) then ω ∈

∞ ∩ n=1

KnN(F).

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