T1) ∅,X?∈τ
∅⊆X∖{a}⇒∅∈P(X∖{a})τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A}|≤ℵ0)}}⇒∅∈τ.
(a∈X⊆X)(|X∖X|=|∅|=0≤ℵ0)⇒X∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}}⇒X∈τ.
T2) A,B∈τ olsun.
I. durum: A,B∈P(X∖{a}) olsun.
A∈P(X∖{a})⇒A⊆X∖{a}B∈P(X∖{a})⇒B⊆X∖{a}}⇒A∩B⊆X∖{a}⇒A∩B∈P(X∖{a})…(1)
τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}…(2)
(1),(2)⇒A∩B∈τ.
II. durum: A,B∈{A|(a∈A⊆X)(|X∖{a}|≤ℵ0)} olsun.
A∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}⇒(a∈A⊆X)(|X∖A|≤ℵ0)B∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}⇒(a∈B⊆X)(|X∖B|≤ℵ0)}⇒
⇒(a∈A∩B⊆X)(|X∖(A∩B)|=|(X∖A)∪(X∖B)|≤ℵ0)
⇒A∩B∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}…(1)
τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}…(2)
(1),(2)⇒A∩B∈τ.
III. durum: A∈P(X∖{a}) ve B∈{A|(a∈A⊆X)(|X∖{a}|≤ℵ0)} olsun.
A∈P(X∖{a})⇒A⊆X∖{a}B∈{A|(a∈A⊆X)(|X∖{a}|≤ℵ0)}⇒(a∈A⊆X)(|X∖{a}|≤ℵ0)}⇒
⇒A∩B⊆X∖{a}⇒A∩B∈P(X∖{a})…(1)
τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}…(2)
(1),(2)⇒A∩B∈τ.
T3) A⊆τ olsun.
A⊆τ
⇒
A⊆P(X∖{a}) ∨ A⊆{A|(a∈A⊆X)(|X∖A}|≤ℵ0)} ∨ A⊆P(X∖{a})∪{A|(a∈A⊆X)(|X∖A}|≤ℵ0)}
I. durum: A⊆P(X∖{a}) olsun.
A∈A⊆P(X∖{a})⇒A⊆X∖{a}⇒⋃A=⋃A∈AA⊆X∖{a}⇒⋃A∈P(X∖{a})τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}}⇒⋃A∈τ.
II. durum: A⊆{A|(a∈A⊆X)(|X∖A}|≤ℵ0)} olsun.
A∈A⊆{A|(a∈A⊆X)(|X∖A}|≤ℵ0)}⇒(a∈A⊆X)(|X∖A}|≤ℵ0)
⇒(a∈⋃A∈AA=⋃A⊆X)(|X∖(⋃A)|=|X∖(⋃A∈AA)|≤|X∖A|≤ℵ0)
⇒⋃A∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}τ=P(X∖{a})∪{A|(a∈A⊆X)(|X∖A|≤ℵ0)}}⇒⋃A∈τ.
III. durum: A⊆P(X∖{a})∪{A|(a∈A⊆X)(|X∖A}|≤ℵ0)} olsun.
A⊆P(X∖{a})∪{A|(a∈A⊆X)(|X∖A}|≤ℵ0)}
⇒
(∃A1⊆2X)(∃A2⊆2X)(A1⊆2X∖{a})(A2⊆{A|(a∈A⊆X)(|X∖A|≤ℵ0)})(A=A1∪A2)
I. ve II. Durum⇒
(⋃A1∈2X∖{a})(⋃A2∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)})(⋃A=(⋃A1)∪(⋃A2))
?⇒
⋃A∈τ.
Not: Soru işaretinin gerekçesi aşağıda verilmiştir.
(A∈2X∖{a})(B∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)})
⇒
A∪B∈{A|(a∈A⊆X)(|X∖A|≤ℵ0)}.