Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider $ ‎\delta = \{ X_n : n \in \mathbb{N} \}$ be a sequence in $[0,1]$ with usual topology.

Does $\delta $ have a subsequence $\{ x‎_{n‎_{k}‎}‎ : n \in \mathbb{N} \}$ that converge to some $x \in [0,1]$?

share|improve this question

1 Answer 1

Yes; this is a consequence of the Bolzano-Weierstrass Theorem, which states more generally that a bounded sequence of real numbers has a convergent subsequence. The fact that the limit lies in $[0, 1]$ is due to the fact that $[0, 1]$ is closed.

One way to prove this is to divide the interval into $[1/2, 1]$ and $[0, 1/2]$, noting that one of these intervals contains infinitely many sequence terms. Subdivide that interval into two subintervals of length $1/4$, and continue.

share|improve this answer
    
the Bolzano- Weierstrass Theorem is that every bounded infinit subset of real numbers has accumulation point. How can we obtain " there is a converge sequence in $[0,1]$ s.t converge to $x \in X$ by the Bolzano- Weierstrass Theorem ? Please help me. –  fatemeh Sep 20 '13 at 8:53
    
So $\delta$ has an accumulation point; try interpreting "accumulation point" in the context of limits. –  T. Bongers Sep 20 '13 at 8:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.