Proof of "Every Cauchy Sequence is Bounded

[fderingoz]

I read the proof of the proposition "every cauchy sequence in a metric spaces is bounded" from

http://www.proofwiki.org/wiki/Every_Cauchy_Sequence_is_Bounded

I don't understand that how we can take m=N_{1} while m>N_{1} ?

In fact i mean that in a metric space (A,d) can we say that

[\forallm,n>N_{1}\Rightarrow d(x_{n},x_{m})<1]\Rightarrow[\foralln\geqN_{1}\Rightarrow d(x_{n},x_{_{N_{1}}})<1]

 

[Erland]

I read the proof of the proposition "every cauchy sequence in a metric spaces is bounded" from

http://www.proofwiki.org/wiki/Every_Cauchy_Sequence_is_Bounded

I don't understand that how we can take m=N_{1} while m>N_{1} ?

In fact i mean that in a metric space (A,d) can we say that

[\forallm,n>N_{1}\Rightarrow d(x_{n},x_{m})<1]\Rightarrow[\foralln\geqN_{1}\Rightarrow d(x_{n},x_{_{N_{1}}})<1]

You are right. This is an error in the wiki. m,n&gt;N should be changed to m,n\ge N wherever it occurs (this also holds with N_1 instead of N). This fits the wiki's definition of Cauchy sequence, which the wiki's proof doesn't.

 

[fderingoz]

Thank you for the answer i also think like you. This is an error in the wiki. But i saw several functional analysis book which write the proof of proposition same as in wiki. So,

Who is wrong?

 

[Erland]

Well, we can define "Cauchy sequence" with either &gt; or \ge, but in the former case, we cannot use N_1 the way it is used in the proof in the wiki. Then we also need an N_2&gt;N_1 to work with, or something like that.