# Closed unit ball in $(C^0[0,1],\mathbb{R})$ equipped with the sup norm is not sequentially compact.

Closed unit ball in $(C^0[0,1],\mathbb{R})$ equipped with the sup norm is not sequentially compact.

I consider $f_n(x) = x^n$ where $0≤x≤1$ Then $f_n(x)$ is clearly a subset of the closed unit ball. Also $f_n(x)$ converges pointwise to $f(x) = 0$ for $0≤x<1$ and $1$ for $x=1$. $f$ is not continuous, so $f_n$ does not converge uniformly to $f$? Then I am stuck?

A subset of a normed space is sequentially compact if every sequence with all terms in the subset has at least one convergent subsequence with its limit in the subset.

• There's something off with the "$<$" and "$\le$" signs. $f_n(x)$ should be $x^n$ for all $0\le x\le 1$ (instead of $0\le x<1$), while $f(x)$ should be $0$ for $0\le x<1$, instead of $0<x\le 1$. – user228113 Jan 29 '16 at 16:27
• Corrected those thanks – Gerniant Jan 29 '16 at 16:29

Suppose there exists a subsequence $f_{n_k}\to g\in C^0[0,1]$ (with $\lVert g\rVert_\infty\le1$) uniformly on $[0,1]$. Then, $f_{n_k}\to g$ pointwise. But we know that $(f_{n})_{n\in\Bbb N}$ converges pointwise to a non-continuous function - and so must $(f_{n_k})_{k\in\Bbb N}$. But $g$ is continuous by definition. Absurd.