Real Analysis Chapter 4 Solutions
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چکیده
Since X is infinite it contains two distinct points x and y. Suppose there exist disjoint open sets A and B (in the cofinite topology) such that x ∈ A and y ∈ B. Then A ⊆ Bc, which is finite so A is also finite. This is a contradiction because Ac is finite but X = A ∪Ac is infinite. Hence, the cofinite topology on X is not T2. Suppose that X is countable, and let x ∈ X. Choose a surjection q : N→ {x}c, and for each n ∈ N set An := {q(k)}k=1 and Bn := A c n, so that x ∈ Bn and Bn is open. If U is open and x ∈ U then U c ⊆ {x}c is finite and hence U c ⊆ AN for some sufficiently large N ∈ N. This implies that BN ⊆ U, so {Bn}n=1 is a neighbourhood base at x. The cofinite topology on X is therefore first countable.
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تاریخ انتشار 2015