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The proof I was given used interval bisection and if I apply that proof to an open interval I don't see where it would go wrong.

Also is it true that the Bolzano Weierstrass property doesn't hold for any open interval?

Thanks!

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This is not a direct answer to your question but it does complement other people's answers.

You need to be clear what statement is:

Bolzano-Weierstrass theorem says every bounded sequence $\{x_n\}_{n\in\mathbb{N}}$ in $\mathbb{R}^n$ has a convergent subsequence in $\mathbb{R}^n$.

Sequential compactness theorem says that every sequence $\{x_n\}_{n\in\mathbb{N}}$ in a closed and bounded subset of $\mathbb{R}^n$ has a convergent subsequence in that set.

Now it is not true that if $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $(0,1)$ then it has a convergent subsequence in the open interval $(0,1)$, as pointed out by Dominic. But it is true that $\{x_n\}_{n\in\mathbb{N}}$ has a convergent subsequence in $\mathbb{R}$ as result of Bolzano-Weierstrass.

So the Bolzano-Weierstrass theorem does not fail for open intervals, but the sequential compactness property does. Since $(0,1)$ is not closed, it does not contain all the limit points. If limit point that the subsequence of $\{x_n\}_{n\in\mathbb{N}}$ that wants to converge to is not in the open interval, like in the example, it does not satisfy sequential compactness property but Bolzano-Weierstrass is still correct.

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Hint: Think about the sequence $$\left\{\frac{1}{n}\right\}_{n=2}^\infty\subset(0,1).$$The problem is that the bisection may put you closer and closer to an endpoint, which isn't contained in an open interval.

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  • $\begingroup$ seems like a popular counterexample :) $\endgroup$ – Dominic Michaelis Apr 5 '13 at 12:06
  • $\begingroup$ Agreed!${}{}{}$ $\endgroup$ – Clayton Apr 5 '13 at 12:08
  • $\begingroup$ Okay I see your point, but then is it the case that any sequence in an open interval converges to some point (not necessarily in the open interval)? $\endgroup$ – user68293 Apr 5 '13 at 12:12
  • $\begingroup$ daochen wang it can't converge to something which is not in the intervall, because it is not there. It is like saying that $a_n=n$ converges to infty, you can always add points to make sequences converge. $\endgroup$ – Dominic Michaelis Apr 5 '13 at 12:13
  • $\begingroup$ @DaochenWang: The sequence will necessarily have a convergent subsequence as long as the interval it is contained in is bounded (this fact due to Bolzano-Weierstrass). However, if the limit isn't in the open interval, it is said to diverge with respect to that particular interval (as above). $\endgroup$ – Clayton Apr 5 '13 at 12:16
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the Bolzano proof uses the lemma that a nested sequence of closed intervals has non-empty intersection.

The same counter-example others have given is esssentially the counter-example for this lemma.

Namely: $$I_n =(0,1/n)$$

is a nested sequence of open intervals whose intersection is empty.

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Because the Intersection would be empty, if you do the same.
When I think of the same proof you do, we need the fact that if $$C_{i+1}\subset C_{i}$$ for all $i$, And $C_i$ are closed intervalls $$\operatorname{diam} C_{i+1} \leq \frac{1}{2} \operatorname{diam} C_i$$ where diam is the length of the intervall, then $$\bigcap_{i=1}^\infty C_i=\{ x_0 \}$$ For open Intervalls this one is not true take $$\bigcap_{i=1}^\infty \left(0,\frac{1}{2^i}\right) =\varnothing$$

For example taking the open Interval $(0,1)$ and the sequence $$a_n=\frac{1}{n}$$ this one is bounded, but it has no limit point in $(0,1)$.

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