Bolzano-wierstrass theorem and its dilemma when applied to cauchy sequence

[jessicaw]

From this theorem, we know that every cauchy sequence is bounded so there exists an accumulation point a$\in \mathbb{R}^k$. This is the limit point of the cauchy seqence.
It shows that there is only one accumulation point as limit point is unique.
However is it a must that there is only one accumulation point predicted by the BW theroem? How to show that there is only one accumulation point in a bounded set?Thanks!

 

[HallsofIvy]

The Bolzano-Wierstrasse theorem only says that a bounded sequence of real numbers must have at least on convergent subsequence. it does NOT follow that there exist only one.

For example, the squence 0, 3/2, -2/3, 5/4, ..., $(-1)^n+ 1/n$, ... is bounded- -2 is a lower bound and 2 is an upper bound. It has two subsequence, one which converges to 1 and one which converges to -1.

And the Bolzano-Weierstrasse theorem does NOT show that every Cauchy sequence is bounded. We can prove that every Cauchy sequence is bounded directly from the definition of "Cauchy sequence". The Bolzano-Weierstrasse theorem then tells us that there exist at least one convergent subsequence. Once we have that limit "in hand", we use the property of a Cauchy sequence to show that it is the only limit.

 

[deluks917]

If we can assume the sequence is Cauchy why do we need Bolzano-
Weierstrass? Don't you just need to show the only limit point is the limit of the sequence?

 

[jessicaw]

If we can assume the sequence is Cauchy why do we need Bolzano-
Weierstrass? Don't you just need to show the only limit point is the limit of the sequence?

You said the only limit(accumulation) point and this is what i am confused about. By Bolzano weieteass theroem, there is only oneaccumulation point in the cauchy sequence?why?

 

[jessicaw]

The Bolzano-Wierstrasse theorem only says that a bounded sequence of real numbers must have at least on convergent subsequence. it does NOT follow that there exist only one.

For example, the squence 0, 3/2, -2/3, 5/4, ..., $(-1)^n+ 1/n$, ... is bounded- -2 is a lower bound and 2 is an upper bound. It has two subsequence, one which converges to 1 and one which converges to -1.

And the Bolzano-Weierstrasse theorem does NOT show that every Cauchy sequence is bounded. We can prove that every Cauchy sequence is bounded directly from the definition of "Cauchy sequence". The Bolzano-Weierstrasse theorem then tells us that there exist at least one convergent subsequence. Once we have that limit "in hand", we use the property of a Cauchy sequence to show that it is the only limit.

you said that there is more than one subsequential limits(accumulation points), how do we know which one is the limit of the cauchy sequence?

 

[deluks917]

The Bolzano-Weierstrass says that every bounded sequence of real numbers has at least one convergent sub-sequence. Cauchy sequences is another name for convergent sequence. A Convergent sequence has one and only one limit point (the point it converges to). We don't need the BW theorem to prove this fact about Cauchy sequences.

you said that there is more than one subsequential limits(accumulation points), how do we know which one is the limit of the cauchy sequence?

There can only be more then one accumulation point if the sequence isn't Cauchy(convergent). Its easy to prove if a_n converges it has only one limit point. Assume lim n->inf a_n = a then a is a limit point. Assume it has another limit point b. The sequence can't converge to a and be arbitrarily close to b. Hence we have a contradiction. a_n can therefore only have one limit point.

 

[jessicaw]

The Bolzano-Weierstrass says that every bounded sequence of real numbers has at least one convergent sub-sequence. Cauchy sequences is another name for convergent sequence. A Convergent sequence has one and only one limit point (the point it converges to). We don't need the BW theorem to prove this fact about Cauchy sequences.



There can only be more then one accumulation point if the sequence isn't Cauchy(convergent). Its easy to prove if a_n converges it has only one limit point. Assume lim n->inf a_n = a then a is a limit point. Assume it has another limit point b. The sequence can't converge to a and be arbitrarily close to b. Hence we have a contradiction. a_n can therefore only have one limit point.

i know there is only 1, but which one?

 

[lgd0612]

From this theorem, we know that every cauchy sequence is bounded so there exists an accumulation point a$\in \mathbb{R}^k$. This is the limit point of the cauchy seqence.
It shows that there is only one accumulation point as limit point is unique.
However is it a must that there is only one accumulation point predicted by the BW theroem? How to show that there is only one accumulation point in a bounded set?


Thanks!

By "It", do you mean the BW thm or the definition of Cauchy sequence?

If you apply BW thm to a cauchy sequence, it only tells you that for this cauchy sequence in R^n, there must exist at least 1 limit point.

It is the definition of Cauchy sequence tells us that there can only be one limit point thus it is unique.