- #1

CoachBryan

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- 0

I really appreciate the help guys. Thanks.

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- Thread starter CoachBryan
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- #1

CoachBryan

- 14

- 0

I really appreciate the help guys. Thanks.

- #2

homeomorphic

- 1,773

- 130

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.

- #3

CoachBryan

- 14

- 0

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).

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