# A bounded sequence has a sub-sequence that converge.

Let $E$ a normed space. We know that if $E$ has finite dimension, then all bounded sequence has a subsequence that converge (Bolzano-Weierstrass theorem). My course says that it's wrong a priori if $E$ has infinite dimension.

My question : If $E$ has infinite dimension and is complete, is the revious result true ? I would say yes since to show Bolzano-weierstrass we use the fact that a Cauchy sequence converge (and thus that the space it is complete). But since I'm not sure, I prefer to ask.

In an infinite dimensional Hilbert space we can take a l.i. set $\{e_1,e_2,\ldots,e_n,\ldots\}$. Since $\|e_j-e_k\|=1$ for any $j\neq k$, this set has no accumulation point.