“…It is not hard to verify, however, that ifẋ ∈ HS and 1 HSẋ ∈Ȧ, then {n | 1 ẋ =ȧn} is finite, and by an easy density argument, 1 does not decide the value of0 ∈ȧn for any finitely many reals. 4 This leads to the sometimes additional requirement that κ is uncountable in the definitions of measurable and strongly compact cardinals. For our purposes, however, it is better to allow ℵ 0 to be considered as measurable or strongly compact.…”