Is the ball compact?

Question

Consider the space $C[0,1]$ of the continuous functions $f\colon [0,1]\to \Bbb R$, with $d_\infty(f,g)= \max_{x\in [0,1]} \lvert f(x)-g(x) \rvert$.

Is the unit ball $\bar B _1 (0)$ compact, where $0$ is the zero constant function?

I just do not understand this question. The unit ball is closed because it is the closure of the ball, I assume, so I need to show boundedness. Can someone please help me understand the question and give me a hint? Also, how can one construct ball around a function?

Thanks in advance!

Answer 1

No. Heine–Borel theorem shows closed and bounded implies compact in $\mathbb{R}^n$, but in general it's not true.

For your question, Riesz lemma told us a unit ball is compact iff the normed vector space is finite dimensional.

If you want to use sequential compact to draw a contradiction, consider $f_n(t)=t^n$, $0\le t\le 1$. Then $\{f_n\} \subset \overline{B(0,1)}$ , but no subsequence of $\{f_n\}$ converges in $C[0,1]$ (with the sup norm).

Answer 2

Hint: Find a sequence in $\bar B_1(0)$ that has no subsequence converging in $C[0,1], d_\infty.$