# Real analysis

1. Nov 15, 2013

### Lee33

1. The problem statement, all variables and given/known data

Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact.

2. The attempt at a solution

I was given a hint, to look at the sequence of continuous functions $f_n(x) = x^n$ on the closed ball in $C([0,1])$. Why is that sequence continuous? Isn't it discontinuous in the closed ball $[0,1]$?

Last edited: Nov 15, 2013
2. Nov 15, 2013

### pasmith

Continuity is not a notion that applies to sequences. Convergence is.

I assume you are using the sup metric, $d(f,g) = \sup \{|f(x) - g(x)| : x \in [0,1]\}$.

The function $f_n : x \mapsto x^n$ is continuous (indeed, differentiable) on $[0,1]$ for any $n \in \mathbb{N}$. Hence $f_n \in C([0,1])$. It is straightforward to show that $d(f_n,0) = 1$, so $f_n$ is in the closed unit ball for all $n$.

The question is: Does this sequence converge to a limit in $C([0,1])$ or not? Does it have any subsequences which do? And what does this tell you about the compactness of the closed unit ball in $C([0,1])$?

3. Nov 15, 2013

### Lee33

But I thought $f_n : x \mapsto x^n$ is not continuous on $[0,1]$ since $f(x) = \begin{cases} 1 & \quad \text{if }x= 1 \\ 0 & \quad \text{if }x\in [0,1) \\ \end{cases}$

4. Nov 15, 2013

### LCKurtz

$f_n$ is a polynomial and is continuous for all $x$. Surely you know that. All polynomials are.

5. Nov 15, 2013

### Lee33

LCKurtz - Wow, you're right! I should go run outside and clear my mind. Very dumb mistake by me. Thank you both!

6. Nov 15, 2013

### WannabeNewton

What you wrote down is the pointwise limit of $(f_n)$ and indeed the pointwise limit is discontinuous. However $f_n$ is continuous for all $n$.

Regardless, does the uniform limit of $(f_n)$ exist? Furthermore, does any subsequence $(f_{n_k})$ of $(f_n)$ have a uniform limit? Recall that convergence in $C([0,1])$ is uniform convergence in $[0,1]$. And if no $(f_{n_k})$ has a uniform limit then what does that tell you about sequential compactness of the closed unit ball in $C([0,1])$?

7. Nov 15, 2013

### Lee33

Thank you, WannabeNewton! I know what to do now. Since it is not sequentially compact, therefore it is not compact.