Footnote 1:

Let T be the circle and let K be the countable subset. By a counting argument, there is a rotation ρ such that for no member x of K do we have an n such that ρn(x)∈K. Let A=K∪ρ(K)∪ρ2(K)∪.... Then (TA)∪A=T and (TA)∪ρ(A)=TK.

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