The sequence {Zn} is convergent
iff.
given any ε>0,
there exists an integer N=N(ε)>0 such that |Zm-Zn|<ε for all m,n>N
以上是定理敘述, 由左到右的證明我OK, 但是由右到左的證明我把課本上的
最後一部分證明改了一點，不知道有沒有寫錯, 請問有沒有好心人願意幫我看
前半段和原始證明一樣
Proof:
Conversely, Suppose the seq. Zn satisfies the Cauchy convergence criterion.
Choosing ε=1, there exists an integer N(1)>0 such that |Zn-Zmo|<1
for all mo>N(1).
Then |Zn|=|Zn-Zmo+Zmo|≦|Zn-Zmo|+|Zmo|<1+|Zmo|
Let M=max{|Z1|, |Z2|,..,|Zmo|,1+|Zmo|}
Then |Zn|≦M.
Thus, {Zn} is a bounded seq..
By the Bolzano-Weierstrass Thm, {Zn} has at least one limit point,
註:極限點的定義
α is a limit point of a seq. {Zn}
if for anyε>0, |Zn-α|<ε.
從這裡開始, 我要證明{Zn}只有唯一一個極限點
Let α and β are two limit points of two subseq.{Zni}, {Zmj} of {Zn}
such that lim Zni=α lim Zmj=β.
Then for any ε>0 there exists an integer N(ε/3)>0 such that
|α-β|=|α-Zni+Zni+Zmj-β-Zmj|≦|α-Zni|+|Zni-Zmj|+|Zmj-β|<ε/3+ε/3+ε/3=ε.
we have α=β.
接著我要證明如果一個數列只有一個極限點, 那此數列為收斂數列.
Claim: If a seq. has only one limit point
then the limit of the seq. exists.
Proof:
Let α be the limit point of {Zn}.
Then for any ε>0 , |Zn-α|<ε, for all integers n.
Thus, for any ε>0 , there is an integer N(ε)>0
such that |Zn-α|<ε, for all n>N(ε).
We have lim Zn=α, that is the seq. converges.
證明完畢
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※ 編輯: rebe212296 來自: 123.193.56.234 (11/15 22:49)
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