# Closed and Bounded but not compact

Let $(C[0,1];d_{\infty})$ the metric space with $C[0,1]$ the continuous functions on $[0,1]$ and $d_{\infty}(f,g)=\max_{x \in[0,1]}|f(x)-g(x)|$

Prove that $$S=\{f:|f(x)| \leq 1\}$$ is closed and bounded but no compact.

My attempt: I've already proved that is bounded using a ball with radius $2$. But I can't figure out how can I prove that is closed and is not compact, I tried with the limit points and the complement but I'm lost.

It is closed because the uniform limit of continuous functions is continuous, and the limit function will belong to S, because it is also the pointwise limit.

For non-compactness, since we are in a metric space, it is enough to find a sequence in S which (you can show) has no convergent subsequence. $\{x^n\}_{n\geq 1}$ will do the trick.

Let $X = C([0, 1])$, with $d(f, g) = \sup_{0 \le x \le 1} |f(x) - g(x)|$. Let $B = \{f \in X: d(f, 0) \le 1\}$. Show that $B$ is closed and bounded, but $B$ is not compact.

1. $B$ is bounded. Set $\|f\| = d(f, 0)$ for $f \in C([0, 1])$. Then $\|f\| \le 1$ for all $f \in B$.
2. $B$ is closed. Let $f \in C([0, 1])$ be a limit point of $B$. Then there exists a Cauchy sequence $\{f_n\}$ such that$$\lim_{n \to \infty} d(f_n, f) = 0.$$Then$$\|f\| \le d(f, f_n) + d(f_n , 0) \le d(f, f_n) + 1$$for all $n$. Letting $n$ go to infinity, we get $\|f\| \le 1$. That is, $f \in B$.
3. $B$ is not compact. For $n \ge 1$, we define$$g_n(x) = \begin{cases} 1 - nx & \text{if }0 \le x \le {1\over{n}} \\ 0 & \text{if }{1\over{n}} \le x \le 1.\end{cases}$$Then $g_n \in B$. Note that $g_n(x)$ is a decreasing function. Let $m < n$. Then$$g_m(x) - g_n(x) = \begin{cases} (n-m)x & \text{if }0 \le x \le {1\over{n}} \\ 1 - mx & \text{if } {1\over{n}} \le x \le {1\over{m}} \\ 0 & \text{ if} {1\over{m}} \le x \le 1.\end{cases}$$It is clear that$$d(g_m, g_n) = {{n-m}\over{n}} \ge {1\over{m+1}}.$$If $m > n$, then$$d(g_m, g_n) \ge {1\over{n+1}} \ge {1\over{m+1}}.$$This implies that $\{g_1, g_2, \dots\}$ is a bounded closed subset of $C([0, 1])$. Let$$U_0 = C([0, 1]) \setminus \{g_1, g_2, \dots\},$$which is clearly an open set. Also, set$$U_m = \left\{f \in C([0, 1]): d(f, g_m) < {1\over{2(m+1)}}\right\}$$for $m \in \mathbb{N}$. Then$$U_m \cap \{g_1, g_2, \dots\} = \{g_m\}$$and $U_m$ is an open set of $C([0, 1])$. It is clear that $\{U_i: i \ge 0\}$ is an open covering of $B$. But any finite number of $U_i$'s can not cover $B$.

For each $n\in \mathbb{N}$, choose a function $f_n:[0,1]\rightarrow[0,1]$ so that $f$ vanishes off of the closed inteerval $[1/(n+1),1/n]$, $0\le f \le 1$, and so that $f = 1$ in the middle of the interval. For all $m$, $n$, $$d_{\infty}(f_m, f_n) = 1$$ if $m\not=n$. This precludes compactness.

Closed and bounded follow from the fact that $S = \{ f | d_\infty(f,0) \le 1 \}$.

Let $\phi(f) = \int_0^{1 \over 2} f(x) dx - \int_{1 \over 2}^1 f(x) dx$; it is easy to verify that $\phi$ is continuous.

If $S$ was compact then $\phi$ would have a maximum on $S$. However, we have $\sup_{f \in S} \phi(f) = 1$, but $\phi(f) < 1$ for all $f \in S$. Hence $S$ is not compact.

It is closed because if you have a function $f \notin S$ then you have for some $a$ that $|f(a)| > 1$. Now this means that any function $g$ such that $d(g,f) < (|f(x)|-1)/2$ is not in $S$ since $|g(a)| > 1$ by triangle inequality. This shows that $\overline S$ is open and therefore $S$ is closed.

It's not compact since the sequence $f_n = F(x-1)$ has no converging subsequence, where $F(x)=2x$ for $0<x<1/2$, $2-2x$ for $1/2<x<1$ and zero elswhere. It cannot since $d(f_j, f_k) = 1$ for every different $j$ and $k$.