This question is taken from Saxe K -Beginning Functional Analysis.

Show that the closed unit ball in $C[0,1]$ is not compact by proving that it is not sequentially compact.

(It's assumed that we are using the uniform norm).

I've been working on this for ages but I could not come up with any sequence $\{f_n\}$ in the unit ball such that there exists $N\in \mathbb{N}$ such that for all $m,n\geq N$ we have that $d(f_n,f_m)>c$. Should be a nice example of this, please help me!

  • $\begingroup$ Was the answer below helpful? $\endgroup$ – JohnD Feb 2 '13 at 22:58

Consider $f_n(t)=t^n$, $0\le t\le 1$. Then $\{f_n\} \subset \overline{B(0,1)}$ (closed unit ball), but no subsequence of $\{f_n\}$ converges in $C[0,1]$ (with the sup norm).

  • $\begingroup$ This maid me think... The sequence given above is uniformly Cauchy, since for every $\epsilon >0$, there exists a $N\in \mathbb{N}$ such that $n,m>N$ implies that $sup_{x\in [0,1]}|f_n(x)-f_m(x)|<\epsilon$, hence it should be uniformly convergent, hence converge to a continuous function... But the limit equals $1$ for $x=1$ and $0$ for $x\in [0,1)$... This is what you meant with no convergent sub sequence right...? How does this make sense???? $\endgroup$ – harajm Feb 3 '13 at 19:06
  • $\begingroup$ @harajm: As you say, the sequence $\{t^n\}$, $0\le t\le 1$, cannot be a uniformly convergent sequence since the uniform limit of a sequence of continuous functions is again continuous. So think about where things break down... $\endgroup$ – JohnD Feb 3 '13 at 20:03
  • $\begingroup$ Could you please explain why it is not uniformly Cauchy???? Because for me, $n,m>N$ implies that $sup_{x\in[0,1]}|f_n(x)−f_m(x)|<\epsilon$. Could you explain why this isn't the case??? $\endgroup$ – harajm Feb 4 '13 at 1:31

Hint: each subsequence should converge uniformly to the pointwise limit, which is not continuous.

So take any bounded sequence in $C[0,1]$ which converges pointwise to a non-continuous function.


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