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.