## § Proof of Heine Borel from Munkres (compact iff closed, bounded)

We wish to show that compact iff closed and bounded in $\mathbb R$.
• Let $S$ be closed and bounded. We wish to show that $S$ is compact. Since $S$ is bounded, $S$ is contained in some large closed interval $I$. (1) Closed intervals are compact in the order topology of a complete total order, and thus $I$ is compact ( $\mathbb R$ is a complete total order). Then (2) $S$ is a closed subset of a compact set $I$, and is thus compact. Hence closed and bounded implies compact.
• Let $S$ be compact. We wish to show that $S$ is closed and bounded. Cover $S$ with open balls of increasing radii. This is an open cover; Extract a finite subcover. From the finite subcover, pick the largest open ball $I$. $S$ is entirely contained in $I$, and is thus bounded. (3) $S$ is closed, as it is a compact subset of a Haussdorff space. Hence, we are done.

#### § (1) Closed intervals are compact in the topology of a complete total order.

Let us work with a complete total order $T$ (in our case $T = \mathbb R$). We equip $T$ with the order topology (which matches the usual topology on $\mathbb R$). Let $[l, r]$ be a closed interval. Let $\{ U_i \}$ be an open cover of $[l, r]$. Let $M$ (for middle) be the set of points $m$ such that $[l, m]$ has a finite cover using $\{ U_i \}$ That is,
$M \equiv \{ m \in [l, r] : [l, m] \text{ has finite cover} \}$
• CLAIM 1: $lub(M) \in M$.
• Pick an open $lub(M) \in V \in \{ U_i \}$.
• As $V$ is open, there is some $(c \in V) < (lub(M) \in V)$.
• if $c \in M$: the finite cover of $[l, c]$ along with $V$ give a finite cover for $lub(M)$. Thus $lub(M) \in M$.
• if $c \not \in M$, then $c < lub(M)$ and $c$ is an upper bound for $M$, which is absurd.
• Thus, $lub(M) \in M$.
• CLAIM 2: $lub(M) = r$. This implies that $r \in M$, and $[l, r]$ has a finite subcover using $U_i$.
• For contradiction of Claim 2, assume that $lub(M) \neq r$.
• Pick some open set $O$ in the cover that contains $lub(M)$: $lub(M) \in O \in \{ U_i \}_i$.
• As $O$ is open, $O$ contains some point $c$ (for contradiction) that is after $lub(M)$: $lub(M) < c$.
• Rewriting: $c \in O \land c > lub(M)$. So, the interval $[l, c]$ has the same cover as $[l, lub(M)]$.
• Hence, $[l, c]$ has a finite cover, thus $c \in M$.
• CONTRADICTION: $c \in M \land c > lub(M)$, which is absurd. We would have $c = lub(M)$.
• Thus, this means that $lub(M) = r$, and thus the entire interval $[l, r]$ has finite subcover.

#### § (2) Closed subset of a compact set is compact

Let $B$ be a closed subset of a compact space $K$. Take an open cover $\{ U_i \}$ of $B$. See that $\{ B^c, U_i \}$ is a cover of the full space $K$, and hence has a finite subcover. This subcover will be of the form $\{B^c, U_j\}$. The $\{U_j\}$ are a finite subcover of $B$. Thus, $B$ is compact as we have extracted a finite subcover of an open cover.

#### § (3) Compact subset of Haussdorf space is closed

Morally, this is true because in a Haussdorff space, single point subsets are closed. Compactness pushes this local property to a global property --- The entire compact set itself becomes closed.
• Let $S$ be a compact subset of a haussdorf space $X$.
• For any point $q \not \in S$, we need to show the existence of an open set $q \in Q$ such that $S \cap Q = \emptyset$.
• For each point $s \in S$, use Haussdorf to find separating sets $s \in O(s; q)$, $q \in O(q; s)$ such that $O(s; q) \cap Q(q; s) = \emptyset$.
• See that the sets $\{ O(s; q) : s \in S \}$ are a cover of $S$.
• Extract a finite subcover of this, say $\{ O(s_i; q) : s \in S \}$.
• Use this finite subcover to separate $q$ from $S$.
• Now, pick the open set $Q \equiv \cap \{ O(q; s_i) \}$, which is open since it's a finite intersection.
• See that this $Q$ separates $q$ from $S$.
• We have that $Q \cap O(s_i, q) = \emptyset$ for each $O(s_i, q)$.
• Thus, $Q \cap (\cup O(s_i; q)) = \emptyset$, and thus $Q \cap S = \emptyset$ as $S \subseteq \cup O(s_i; q)$.
• if $q \not \in S$, we have an open $Q$ that separates $q$ from $S$, thus $q$ is NOT A LIMIT --- not every open nbhd of $Q$ has non-empty intersection with $S$.
• Contrapositive: All limit points of $S$ are in $S$. Thus, $S$ is closed. (4)

#### § (4) A set with all limit points is closed (complement of open)

Let $S$ be a set that has all its limit points. Consider the complement set $T$. We will show that $T$ is open.
• all points in $T$: have open that separates them from $S$. Union of all of these opens is $T$. $T$ open: infinite union of opens. $S$: the complement of an open set, closed.
More elaborately:
• for all all $t \in T$, $t \not in S$ (by defn).
• $t$ is not a limit point of $S$ ( $S$ has all limit points).
• Thus, there is an open $U_t$ such that $t \in U_t$ and $U_t \cap S = \emptyset$.
• Define: $T' \equiv \cup_{t \in T} U_t$. Claim: $T' = T$.
• As $U_t$ contains no elements of $S$, $T'$ contains no element of $S$.
• As $U_t$ contains $t$, $T'$ contains all $t$.
• Thus $T'$ contains all $t \in T$, and no element of $S$. So $T'$ is a complement of $S$. $T' = T$.
• $T$ is a infinite union of opens. Thus $T$ is open.
• $S$ is complement of open set $T$. $S$ is closed.