Understanding proof that every locally compact subset of $C[0,1]$ is nowhere dense

Question

This is Problem 1.17 from Billingsley's Convergence of Probability Measures.

Let $C=C[0,1]$ be the space of continuous functions with the sup norm. Show that every locally compact subset of $C$ is nowhere dense.

Below is the solution to this problem, but I have several things I don't understand. Since we assume the contrary, we have some open ball $B \subset \bar{L}$.

However, how do we arrange that $\bar{B} \cap L$ is compact? And how do we choose points from $B$ so that $\rho(x_m,x_n)\ge \epsilon>0$ for $m\neq n$? Also, we can choose such $y_n$ from $\bar{B}$, but how do we ensure they are also from $L$? Finally, how does this conclude that $\bar{B} \cap L$ is not compact?

I would greatly appreciate some help.

Answer 1

We can arrange that $\overline{B}\cap L$ is compact because $L$ is locally compact. Since $B\cap L$ is an open neighbourhoods of some point $x$ in $\overline{B}\cap L$, there is a relatively compact neighbourhood $U$ of $x$ with $U\subseteq B\cap L$. By making $B$ smaller if necessary, we can ensure that $\overline{B}\cap L\subseteq \overline{U}$, which is compact. (Note that $U$ and $\overline{U}$ are subsets of $L$ here.) We are allowed to consider a smaller ball because we need only show that $\overline{L}$ has empty interior, and we are working with arbitrary balls anyway.

To find the points $x_n$, we must use a fact about the space $C[0,1]$: it is an infinite dimensional Banach space. Here we can invoke the Riesz geometric lemma to get the points we are after.

To find the points from $L$ that are close to the points $x_n$, you need to use the fact that $L$ is dense in $B$.

Finally, once we have chosen our sequence $(y_n)$, we note that this sequence cannot possibly have a convergent subsequence because we have $\rho(y_n, y_m) \geq \varepsilon/3$ for all $m,n\in\Bbb N$ with $m\neq n$. This means the sequence $(y_n)$ does not satisfy the Cauchy condition and so cannot be convergent, and every sequence in a compact metric space must have convergent subsequence.

Answer 2

1.- Since $L$ is locally compact and $\overline{B}$ is closed and bounded you immediately obtain that $\overline{B}\cap L$ is compact in $C([0,1])$. This is just the definition of being locally compact (the main point here is that $\overline{B}$ is bounded).

2.- Now that we know that $\overline{B}\cap L$ is compact, let's consider any $\varepsilon>0$ fixed. Since $C([0,1])$ is an infinite-dimensional Banach space we deduce the existence of a sequence $\{x_n\}_{n\in\mathbb{N}}\subseteq C([0,1])$ satisfying $$ \forall n\in\mathbb{N}, \quad \Vert x_n\Vert_{C([0,1])}\leq 1 \quad \wedge \quad \forall n\neq m \Vert x_n-x_m\Vert_{C([0,1])}\geq \varepsilon. $$ This is a consequence of Riesz's Theorem which says that for any infinite-dimensional Banach space such a sequence always exists. Note that, since $\Vert x_n\Vert_{C([0,1])}\leq 1$ for every $n\in\mathbb{N}$, we deduce that $\{x_n\}_{n\in\mathbb{N}}\subseteq \overline{B}$.

3.- Finally, by hypothesis, we know that $\overline{B}\cap L$ is dense in $\overline{B}$ and hence, by density, for every $n\in\mathbb{N}$ we can find an element $y_n\in\overline{B}\cap L$ such that $$ \Vert x_n-y_n\Vert\leq \tfrac{\varepsilon}{3}, \quad \forall n\in\mathbb{N}. $$ Thus, the sequence $\{y_n\}_{n\in\mathbb{N}}$ satisfy \begin{align} \forall n\neq m, \quad \Vert y_n-y_m\Vert_{C([0,1])}&=\Vert x_n-x_m+y_n-x_n+y_m-x_m\Vert_{C([0,1])} \\ & \geq\Vert x_n-x_m\Vert_{C([0,1])}-\Vert y_n-x_n\Vert_{C([0,1])}-\Vert y_m-x_m\Vert_{C([0,1])} \\ & \geq \varepsilon-\tfrac{\varepsilon}{3}-\tfrac{\varepsilon}{3}=\tfrac{\varepsilon}{3}. \end{align} Thus $\{y_n\}_{n\in\mathbb{N}}\subseteq\overline{B}\cap L$ is a bounded sequence which cannot have any convergent subsequence, and hence $\overline{B}\cap L$ cannot be compact.

Edit: I saw you ask something in the comments of SamM's answer, and the answer is yes. This is a (easy to prove) general property: If you have any Banach space $X$, a dense subset $Y\subset X$ and an open set $Z\subseteq X$, then $Y\cap Z$ is dense in $Y$. Thus, since in your case $L$ is dense, $B\cap L$ is dense in $B$.