- Theorem: closed bounded subset of $\R^n$ is compact
- We will prove it for $\R$ and leave the obvious generalization to the reader.
- Key idea: recall that for metric spaces, compactness and sequential compactness are equivalent, so the proof must follow some ideas from Bolzano Weirstrass (sequence in closed bounded set has convergent subsequence).
- Recall that
*that*proof goes by bisection, so let's try to bisect some stuff! - Also recall why this fails in infinite dimensions: you can bisect repeatedly in "all directions" and get volume (measure) to zero, without actually controlling the cardinality. There is no theorem that says "measure 0 = single point". So, the proof must rely on finite dimension and "trapping" a point.
- Take an interval, say $[0, 1]$ and take a cover $\mathcal C$. We want to extract a finite subcover.
- For now, suppose that the cover is made up only of open balls $B(x, \epsilon)$. We can always reduce a cover to a cover of open balls --- For each point $p \in X$ which is covered by $U_p$, take an open ball $B_p \equiv B(p, \epsilon_p) \subseteq U$. A finite subcover of the open balls $\{ B_p \}$ tells us which $U_p$ to pick from the original cover.
- Thus, we shall now assume that $C$ is only made up of epsilon balls of the form $C \equiv \{ B(p, \epsilon_p) \}$.
- If $C$ has a finite subcover, we are done.
- Suppose $C$ has no finite subcover. We will show that this leads to a contradiction.
- Since we have no finite subcover, it must be the case that at $I_0$, there are an infinite number of balls $\{ B \}$. Call this cover of infinite balls $C_0$.
- Now, let the interval $I_1$ be whichever of $[0, 1/2]$ or $[1/2, 1]$ that has infinitely many balls from $C_0$. One of the two intervals must have infinite many balls from $C_0$, for otherwise $C_0$ would be finite, a contradiction. Let $C_1$ be the cover of $I_1$ by taking balls from $C_0$ that lie in $I_1$.
- Repeat the above for $I_1$. This gives us a sequence of nested intervals $\dots \subset I_2 \subset I_1 \subset I_0$, as well as nested covers $\dots \subset C_2 \subset C_1 \subset C_0$.
- For each $i$, pick any epsilon ball $B_i(p_i, \epsilon_i) \in C_i$. This gives us a sequence of centers of balls $\{ p_i \}$. These centers must have a coverging subsequence $\{ q_i \}$ (by bolzano weirstrass) which converges to a limit point $L$.
- Take the ball $B_L \equiv (L, \epsilon_L) \in C$ which covers the limit point $L$.
- Since the sequence $\{ q_i \}$ is cauchy, for $\epsilon_L$, there must exist a natural $N$ such that for all $n \geq N$, the points $\{ q_n : n \geq N \} \subseteq B_L$.
- Thus, we only have finitely many points, $q_{\leq n}$ to cover. Cover each of these by their own ball.
- We have thus successfully found a covering for the full sequence!