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:
(1)
\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:
(2)
\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:
(3)
\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.