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Take the sequence $f_n(t)=t^n$, $0\le t\le 1$. Then $\{f_n\} \subset \overline{B(0,1)}$, but we have no subsequence of $\{f_n\}$ converging in $C([0,1])$. So the unit ball is not compact in $C([0,1])$?

Could someone elaborate on this? Why do we have no subsequence of $\{f_n\}$ converging in $C([0,1])$? What does such a subsequence look like? Where does the subsequence converge if it does not converge in $C([0,1])$?

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    $\begingroup$ What is the metric on $C([0,1])$? $\endgroup$
    – Matias Heikkilä
    May 24, 2016 at 8:23
  • $\begingroup$ @MatiasHeikkilä It has to be the $\|\cdot\|_{\infty} = \sup_{x \in [0,1]}$ $\endgroup$
    – C.S.
    May 24, 2016 at 8:32
  • $\begingroup$ The supremum norm. $\endgroup$
    – csss
    May 24, 2016 at 8:32
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    $\begingroup$ Ah, ok. Then I think you should compare what happens at $t < 1$ and $t=1$. $\endgroup$
    – Matias Heikkilä
    May 24, 2016 at 8:34

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  • Note $$f(x) = \lim_{x \to \infty} f_{n}(x) = \left\{\begin{array}{cc} 0, & 0\leq x < 1 \\ 1, & x =1 \end{array}\right.$$ Can you see now why $f \not \in C[0,1]$
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