Examples of Closed Unit Balls That are NOT Compact

# Examples of Closed Unit Balls That are NOT Compact

Recall from the Closed Unit Ball Criterion for Finite Dimensional Normed Linear Spaces page that if $(X, \| \cdot \|_X)$ is a normed linear space then the closed unit ball of $X$ is compact if and only if $X$ is finite-dimensional.

We will now look at some examples of normed linear spaces whose closed unit balls are not compact. By the theorem referenced above, these spaces will be infinite-dimensional.

Proposition 1: The closed unit ball in $C[0, 1]$ is not compact. |

*In metric spaces, a space is compact if and only if every sequence in the space has a subsequence which converges in the space. This is known as the Bolzano-Weierstrass property.*

**Proof:**$C[0, 1]$ is infinite-dimensional and hence the closed unit ball in $C[0, 1]$ is not compact. Let $\overline{B}(0, 1)$ denote the closed unit ball in $C[0, 1]$. Then:

\begin{align} \quad \overline{B}(0, 1) = \{ f \in C[0, 1] : \| f \|_{\sup} \leq 1 \} \end{align}

- For each $n \in \mathbb{N}$ let $f_n : [0, 1] \to \mathbb{R}$ be defined by $f_n(x) = x^n$. Observe that for each $n \in \mathbb{N}$ we have that:

\begin{align} \quad \| f_n \|_{\sup} = \sup_{x \in [0, 1]} |x^n| = 1 \end{align}

- Then $(f_n)$ is a sequence in $B(0, 1)$. However, observe that $(f_n)$ does not converge uniformly. In fact, $(f_n)$ converges pointwise to:

\begin{align} \quad f(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: 0 \leq x < 1\\ 1 & \mathrm{if} \: x = 1 \end{matrix}\right. \end{align}

- In particular, no subsequence of $(f_n)$ converges uniformly on $[0, 1]$. Now in particular, we have that no subsequence of $(f_n)$ converges in $\bar{B}(0, 1)$ and so $\bar{B}(0, 1)$ is not compact.