Is it true that every Banach space has Bolzano Weierstrass property?

Question

Is it true that every bounded sequence in $C[0,1]$ with sup norm has convergent subsequence? I really feel this statement is true and the crux of the proof will lie in that $C[0,1]$ is complete. But I am unable to prove it mathematically by constructing a Cauchy subsequence for any random sequence ( which will eventually be convergent). Also is my observation correct that every Banach space has this property?

Answer 1

It is not true. The sequence $f_n(x)=\sin nx$ is a uniformly bounded sequence in $C([0,1])$, but it doesn't even have a pointwise convergent subsequence, let alone one in the sup norm.

To identify families where you have a positive result, you can use the Arzela-Ascoli theorem. That is, you just need to check that your family is pointwise bounded and equicontinuous.

Answer 2

Its not true. Consider $f_k(x) = e^{i2\pi kx}$. If $k< j$ are integers, then $$\|f_k-f_j\|_{\infty} = \sup_{x\in[0,1]} |1 - e^{i2\pi (j-k)x}| \ge |1 - e^{i2\pi (j-k)\frac{1}{2(j-k)}}| = 2.$$ Consequently there is no Cauchy subsequence.

Essentially the same example by thinking of Fourier coefficients - you can try the Banach spaces $\ell^p(\mathbb N)$ where now the sequence of elements $f_k \in \ell^p(\mathbb N)$ are defined by $f_k(n) = \delta_{nk}$.