# Prove/Disprove the closure of open ball is complete/ totally bounded/ compact

Consider the metric spaces of all continuous functions on [0,1] mapping into R. The metric is given by $$d(f,g)=\int_0^1|f(t)-g(t)|dt.$$ Prove of disprove the following statements directly.

(i) $B\overline(0,1)$ is complete.

(ii) $B\overline(0,1)$ is totally bounded.

(iii) $B\overline (0,1)$ is compact.

(iv) every sequence in $B\overline (0,1)$ has a subsequence that converges in $B\overline (0,1)$.

I am a freshman of this real analysis course and really confused. Thanks to anyone who is coming for help!

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Hints

(i) Prove that functions $$f_n(t)=\begin{cases} 0&t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\ \frac{1}{2}+n\left(t-\frac{1}{2}\right) &t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\ 1 &t\in\left(\frac{1}{2}+\frac{1}{2n},1\right) \end{cases}$$ forms non-convergent Cauchy sequence.

(iv) Consider "peak" functions $$g_n(t)=\begin{cases} 4^{n+1}\left(t-\frac{1}{2^{n+1}}\right)&t\in\left[\frac{1}{2^{n+1}},\frac{3}{2^{n+1}}\right]\\ -4^{n+1}\left(t-\frac{1}{2^{n}}\right) &t\in\left[\frac{3}{2^{n+2}},\frac{1}{2^n}\right]\\ 0 &t\in[0,1]\setminus\left[\frac{1}{2^{n+1}},\frac{1}{2^{n}}\right] \end{cases}$$ Show that $d(g_m,g_m)\geq 1$ for $m\neq n$. Conclude that $\{g_n:n\in\mathbb{N}\}$ have no convergent subsequence.

(ii) Show that balls $B(g_n, 1/4)$ are contained in $B(0,1)$. Show that there is no finite $1/4$-net to cover this balls, and the consequence there is no finite $1/4$-net for $B(0,1)$.

(iii) Consider the following open cover $$\{B(g_n,1/4):n\in\mathbb{N}\}\cup\left\{B(0,1)\setminus\bigcup\limits_{n=1}^{\infty}\overline{B(g_n,1/8)}\right\}$$ of $\overline{B(0,1)}$. Show there is no finite subcover.

P.S.

There are much more short ways to answer your questions but they are indirect.

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Thank you! But I still have some silly questions...how does the metric d(f,g) measures the space...how to use it? Can you explain to me? Thank you! – Viola Sep 19 '12 at 21:10
Your aim is to prove everything in terms of distance. Example of this way of solving you can find here – Norbert Sep 20 '12 at 4:20