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

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?

• Actually, not at all. The only normed vector spaces whose unit balls are compact are finite-dimensional spaces. – Mindlack Sep 4 '19 at 12:54
• means if X is finite dimensional then it will have BW property? – Believer Sep 4 '19 at 13:50
• Any normed space $X$ has the BW property iff it is finite-dimensional. – Mindlack Sep 4 '19 at 16:05

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.
• I think $f_{n}=x^{n}$ will also work na! – Believer Sep 4 '19 at 13:48
• @believer indeed, this follows from a finite dimensional space being isomorphic to $\mathbb{R}^n$ for some $n$. See math.stackexchange.com/questions/144464/… for details. – cmk Sep 4 '19 at 14:12
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 $$f_k(n) = \delta_{nk}$$.