现在只要证:
∀x0∈A‾ ⟹ x0∈A \forall x_0\in\overline{A}\implies x_0\in A ∀x0∈A⟹x0∈A
根据闭包的定义,∀ϵ>0,B(x0,ϵ)∩A≠∅\forall \epsilon>0,B(x_0,\epsilon)\cap A\neq\varnothing∀ϵ>0,B(x0,ϵ)∩A=∅ ,即:
d(xϵ,x0)<ϵ d(x_\epsilon,x_0)<\epsilon d(xϵ,x0)<ϵ
令 ϵ=1n\epsilon=\frac{1}{n}ϵ=n1 则:
d(xn,x0)<1n d(x_n,x_0)<\frac{1}{n} d(xn,x0)<n1