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I'm trying to make sure I have correctly understood Bolzano-Weierstrass, which states that every compact subset of $\mathbb{R}^{n}$ is sequentially compact, which means that if $A$ is a compact subset of $\mathbb{R}^{n}$, then every sequence in $A$ must have a convergent subsequence (and hence a cluster point).

Consider the sequence $ x_{n} = (\pi*n) \ mod \ 2$. This is a sequence in the compact interval $[0,2]$. By Bolzano-Weierstrass, this sequence must have a cluster point. What is one such cluster point?

I made this sequence myself to test Bolzano-Weierstrass. My understanding of the theorem may be incorrect, and hence no cluster point may exist. Please enlighten me as to which is the case.

EDIT: Just to clarify, my textbook defines a cluster point of a sequence $x_{n}$ to be $x$ such that for all $ \epsilon > 0 $ and $ N $ there is an index $ n > N $ for which $ |x_{n}-x| < \epsilon $

user308485
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It turns out that the sequence $x_n=n\alpha\pmod{1}$ is dense in $[0,1]$, for any irrational $\alpha$. Thus every point in $[0,1]$ (and in fact, every point of $[0,2]$) is a cluster point.

Stella Biderman
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  • How can this be proved? Why is every point in $[0,1]$ a cluster point of the sequence? – user308485 Feb 06 '16 at 06:52
  • Every point is a cluster point because it follows from density that every open interval contains infinitely many points of the sequence. A proof of density can be found here: http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset – Stella Biderman Feb 06 '16 at 07:01
  • @user308485 do you have further questions? If this addresses your concerns, you should consider upcoming or accepting the answer – Stella Biderman Feb 14 '16 at 00:14