Every bounded sequence is Cauchy?

[CoachBryan]

I've been very confused with this proof, because if a sequence { 1, 1, 1, 1, ...} is convergent and bounded by 1, would this be considered to be a Cauchy sequence? I'm wondering if this has an accumulation point as well, by using the Bolzanno-Weirstrauss theorem.

I really appreciate the help guys. Thanks.

 

[homeomorphic]

Cauchy is the same thing as convergent in the real numbers (or any complete metric space). A sequence can be bounded but not converge, like 0, 1, 0, 1, 0, 1...

However, the Bolzano-Weierstrass theorem says that there is a convergent subsequence, which in this case is easy to find directly: 0,0,0,0,0... or 1,1,1,1,1...

You get those by either skipping all the 1's or skipping all the 0's. The full sequence 0, 1, 0, 1, 0, 1... therefore has the accumulation points 0 and 1, since it has subsequences that converge to those points.

1, 1, 1, 1 has the accumulation point 1, the thing that it converges to.

 

[CoachBryan]

Cauchy is the same thing as convergent in the real numbers (or any complete metric space). A sequence can be bounded but not converge, like 0, 1, 0, 1, 0, 1...

However, the Bolzano-Weierstrass theorem says that there is a convergent subsequence, which in this case is easy to find directly: 0,0,0,0,0... or 1,1,1,1,1...

You get those by either skipping all the 1's or skipping all the 0's. The full sequence 0, 1, 0, 1, 0, 1... therefore has the accumulation points 0 and 1, since it has subsequences that converge to those points.

1, 1, 1, 1 has the accumulation point 1, the thing that it converges to.

Gotcha.

Thanks a lot, now it makes sense. So, not every bounded sequence is cauchy, but it's subsequences can be convergent (cauchy).