Solution

Use the isolating sets from the preceding part and make the finite disjointification

B_{n} = A_{n} ∖ ⋃_{j < n} A_{j} .

For

j < n

A_{j} \in / U_{n}

, so

A_{j}^{c} \in U_{n}

. Therefore

B_{n} = A_{n} \cap ⋂_{j < n} A_{j}^{c} \in U_{n}

by finite-intersection closure of the ultrafilter. If

m \neq = n

B_{n} \subseteq A_{n}

and

A_{n} \in / U_{m}

imply

B_{n} \in / U_{m}

. Also

B_{n}

avoids every earlier

A_{j}

, hence every earlier

B_{j}

. Thus

B_{n} \cap B_{m} = \emptyset (n \neq = m), B_{n} \in U_{m} ⟺ m = n .

Only finite intersections were used; no countable-intersection closure of an ultrafilter is being assumed.

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