§ Total Boundedness in a metric space

• A set $A$ is totally bounded iff for any $\epsilon$, there exists a finite $\epsilon$ net $N_\epsilon$.

§ Totally bounded implies bounded

• Let $S$ be a totally bounded set. We want to establish a uniform bound $M$.
• Pick some $\epsilon$. We then get a finite number of points $N$ such that any point in $x$ is $\epsilon$ away from $N$.
• Any two points in $N$ can be at most $N\epsilon$ apart.
• Thus the total distance between any two points $s, t \in S$.
• Let $s, t$ be closest to points $n, n' \in N$.
• Thus $d(s, t) \leq d(s, n) + d(n, n') + d(n', t)$ all of which is bounded by $\epsilon + N\epsilon + \epsilon$ which is less than $(N + 3) \epsilon$.
• Thus we have established a bound of $(N + 3) \epsilon$.

§ For $\mathbb R$, bounded implies totally bounded

• Say set $S$ is bounded by distance $M$.
• Trap $S$ inside an interval of side length $M$, WLOG suppose interval is from $[0, M]$.
• For any adverserial $\epsilon$, pick points at $[0, \epsilon, 2\epsilon, \dots, M]$. These points are the epsilon net which contain $M$: $S \subseteq M \subseteq N_\epsilon$
• The net only needs $M/\epsilon$ points which is finite.
• See that this holds more generally for any $\mathbb R^n$ where we trap in a hypercube and sprinkle points.

§ In infinite dimensions, bounded need not be totally bounded.

• Classic example, sphere in $l^2$. It's clearly bounded by a constant $2$.
• we claim this is not totally bounded.
• Note that any two vectors $e_i, e_j$ have distance $\sqrt(2)$.
• Note that we have an infinite number of basis vectors $e_1, e_2, \dots$.
• Suppose it is totally bounded. Pick $\epsilon = \sqrt(2)/9999$ as an adversary. We get a finite neighborhood set $N_1$.
• Thus, some point in $n \in N_1$ must have an infinite number of basis vectors trapped in it.
• so there must be two basis vectors in it, $e_i, e_j$ such that $e_i \neq e_j$.
• We must have that $d(e_i, e_j) < d(e_i, n) + d(n, e_j)$ by triangle inequality.
• This gives us $\sqrt(2) < \sqrt(2)/9999 + \sqrt(2) / 9999$ which is a contradiction.
• Thus, the sphere is not totally bounded.

§ compact => closed + totally bounded in infinite dim also

• Let $S$ be a compact set.
• it is closed by the usual argument.
• We claim $S$ is totally bounded.
• Let adversary pick $\epsilon$.
• We must establish a finite number of points $N_\epsilon$ such that any point in $S$is in an $\epsilon$ neighbourhood of some $n \in N_\epsilon$.
• Reread that. that's literally compactness.
• Pick an open cover $O$ consisting of an $\epsilon$ ball for each point $s \in S$.
• Extract a finite subcover from this.
• This finite subcover is the finite $\epsilon$ net.
• Thus done. Compact is totally bounded.

§ closed + totally bounded => sequentially compact in infinite dim also

• Let $S$ be a closed and totally bounded set.
• We wish to show that it is sequentially compact.
• Let $x_i$ be a sequence in $S$.
• Perform the classical Bonzalo Weirstrass bisection construction.
• Since $S$ is totally bounded, we can pick any $\epsilon$ and get a finite epsilon net.
• We claim that a closed subset of a totally bounded set is also totally bounded.

def mk_cauchy_sequence(S):
  k = 1
  while True:
    s = choose(S)
    yield s
    Nk = finite_epsilon_net(S=S, epsilon=1/2^k)
    # n ∈ Nk such that n has an infinite number of points from $S$ in
    # the epsilon ball around $n$.
    n = hilbert-epsilon-choose(|{ n ∈ N : S ∩ Ball(n, epsilon) }| = infty)
    S = Ball(n, epsilon) # restrict to ball that has the inif
    k += 1