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

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 $x \in [0,1]$?

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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.
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. – user61527 Sep 20 '13 at 8:55