Proof that $\overline{B}_1(0)$ is not compact in $(C[0,1],||\cdot||_{\infty})$

Question

I am asked to prove the following statement

The set $\overline{B}_1(0)$ is not compact in $(C[0,1],||\cdot||_{\infty})$ and $\overline{B}_1(0)=\{f\in C[0,1]:||f||_{\infty}\leq1\}$

My attempt was the following:

Find a function $h_k(f)$ where $f(x) = ?$ s.t.: $$||h_k(f)-h_m(f)||_{\infty} = const.$$

I am trying to use the definition that a set is called compact if every sequence does have a convergent subsequence.

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The idea behind the constant value came to me when solving the exact same question but for $(\mathbb{R},||\cdot||_{\infty})$, where I found the function $$ f_k(x)= \begin{cases} 1,& \text{if } x= \frac{1}{k}\\ 0, & \text{otherwise} \end{cases} $$ when applying the cauchy-criteria one can define (for example) $\epsilon := \frac{1}{2}$ and see that $||f_k-f_m||_{\infty} = 1 > \epsilon \implies \overline{B}_1(0)$ not compact in $(\mathbb{R},||\cdot||_{\infty})$

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Coming now back to my actual question: Could anyone give me a push in the right direction, what the best way would be to define my $h_k(f)$ or maybe tell me if I am on the complete wrong track with my idea?

It's been some weeks since we introduced the $C[0,1]$ but I am still struggling a bit to work with it thus having a hard time solving this particular excercise.

Thank you in advance :)

P.S.: Since I don't know if $C[0,1]$ is a general used definition: We defined $C[0,1]$ as the space of continuous functions

Answer 1

For each $n\in\mathbb{N}$, define a function $f_n\in C\bigl([0,1]\bigr)$ such that:

• $x\notin\left[\frac1{n+1},\frac1n\right]\implies f_n(x)=0$;
• $(\forall x\in[0,1]):-1\leqslant f_n(x)\leqslant 1$;
• $f_n(x_n)=1$ for some $x_n\in\left[\frac1{n+1},\frac1n\right]$.

It follows from this that $m\neq n\implies\|f_m-f_n\|_\infty=1$. Therefore, $(f_n)_{n\in\mathbb N}$ has no Cauchy subsequence, and therefore no convergent subsequence.