证明：任何有界的复数列必有一个收敛的子数列.

Question

证明：任何有界的复数列必有一个收敛的子数列.

Answer 1

1.
设有界的复数列{z(n)=a(n)+ib(n)}n∈N,
|a(n)|≤|z(n)|≤M==>
{a(n)}n∈N为有界的实数列,则必有一个收敛的子数列
{a(u(k))}k∈N,且Lim{k→∞}a(u(k))=a.
|b(u(k))|≤|z((u(k))|≤M==>
{b(u(k))}k∈N为有界的实数列,则必有一个收敛的子数列
{b(u(v(s)))}s∈N,且Lim{s→∞}b(u(v(s)))=b.
2.
{z(u(v(s)))=a(u(v(s)))+ib(u(v(s)))}s∈N
为{z(n)=a(n)+ib(n)}n∈N的子列,且
Lim{s→∞}z(u(v(s)))=a+ib.