Bounded sequences and convergent subsequences in metric spaces

1. Aug 5, 2011

AxiomOfChoice

Suppose we're in a general normed space, and we're considering a sequence $\{x_n\}$ which is bounded in norm: $\|x_n\| \leq M$ for some $M > 0$. Do we know that $\{x_n\}$ has a convergent subsequence? Why or why not?

I know this is true in $\mathbb R^n$, but is it true in an arbitrary normed space? In particular, since it's true in $\mathbb R$, we know that $\{\|x_n\|\}$ has a convergent subsequence $\{\|x_{n_k}\|\}$ that converges to some $z\in \mathbb R$. My first instinct would be to try to apply the triangle inequality to show that $\|x_{n_k} - x\| \to 0$ for $\|x\| = z$, but the triangle inequality doesn't give me what I want here, since I need to bound $\|x_{n_k} - x\|$ by something that can be made arbitrarily small, but I only have $| \|x_{n_k}\| - \|x\| | \leq \|x_{n_k} - x\|$.

2. Aug 5, 2011

disregardthat

Take the space Q^n over Q, and you will see that it is not the case. E.g. in Q, take an increasing sequence converging to sqrt(2) in R. Any subsequence will converge to sqrt(2) in R, so it will not converge to any element of Q.

3. Aug 5, 2011

mathman

The answer is trivially no. Example in Hilbert Space:
Let xn=(0,0,...,1,0,0....) where the 1 is the nth coordinate. Then ||xn||=1, but the sequence has no convergent subsequences.

The underlying difference is that bounded sets in finite dimensional spaces have a compact closure, but not necessarily in infinite dimensional spaces.

Last edited: Aug 5, 2011
4. Aug 5, 2011

HallsofIvy

Yet another example: $a_1= 3$, $a_2= 3.1$, $a_3= 3.14$, etc. where each term is one more decimal place in the decimal expansion of $\pi$. That is a bounded sequence of rational numbers. Thinking of it as a sequence in the real numbers, it converges to $\pi$ which means every subsequence converges to $\pi$. Thinking of it as a sequence in the rational numbers, clearly no subsequence can converge to a rational number.

More generally, a sequence of rational numbers that converges to an irrational number, cannot have any subsequence that converges to a rational number. Thus, thinking of the sequence as in the rational numbers, no subsequence can converge.

5. Aug 5, 2011

disregardthat

Halls that is more or less the same example as what I brought up.

6. Aug 6, 2011

AxiomOfChoice

Thanks a lot, guys. I should have thought of the "orthonormal sequence in an infinite dimensional Hilbert space" example. Oh well...now I know (hopefully)!