Compactness

Definition. A topological space is compact (C) if every open cover has a finite subcover. It is Lindelöf (L) if every open cover has a countable subcover. It is countably compact (CC) if every countable open cover has a finite subcover.

Proposition 1. $\:$ C $\iff$ (CC and L)

Proof. Evident. $\quad\square$

Definition. A topological space is FIP if every family of closed sets with the finite intersection property has a nonempty intersection.

Proposition 2. $\:$ C $\iff$ FIP

Proof. Any $\{F_i\}$ family of closed sets is dual to $\{V_i\}$ by $V_i = F_i^C$. $\{V_i\}$ is a cover iff $\bigcap F_i = \emptyset$ and $\{V_i\}$ has a finite subcover iff $\{F_i\}$ does not have the finite intersection property. $\quad\square$

Definition. $X$ is CM if for any $Y$, the projection $\pi_2 : X \times Y \to Y$ is a closed map.

Proposition 3. $\:$ C $\iff$ CM

Proof. ($\implies$) Let $F \subseteq X \times Y$ closed and $y \notin \pi_2(F)$. Then $F^C$ contains the tube $X \times \{y\}$. By the tube lemma, there is some neighborhood $V$ of $y$ such that $X \times V \subseteq F^C$, and $V \cap \pi_2(F) = \emptyset$. So $\pi_2(F)^C$ is open. ($\impliedby$) If $X$ not compact, take $\{U_n\}$ a countable cover with no finite subcover. Let $Y = \{0\} \cup \{1/n : n \in \N\} \subseteq R$. Let $V_n = Y \cap [0, 1/n]$ so that $\{V_n\}$ is a countable basis for 0 and $V = \bigcup_{n \in \N} (U_n \times V_n) \subseteq X \times Y$, is open. Choose $A = \pi_2(V^C)$. Clearly $0 \notin A$ yet any neighborhood $W$ of 0 contains $\{1/n : n \ge N\}$. If $W \cap A = \emptyset$, then $\bigcup_{n=1}^N U_n$ would be a cover of $X$, so $0 \in \overline{A}$. $\quad\square$

Definition. A directed set $I$ is any set with a preorder (partial order without antisymmetry) such that any finite set has an upper bound (in particular, for $a,b \in I$, there is some $c \in I$ with $a \le c$ and $b \le c$). A net $\{x_i\}_{i \in I}$ is any family indexed by a directed set. A subnet $\{x_{f(i)}\}$ is a net defined by $f : J \to I$ which is order-preserving and cofinal (for any $i \in I$, there’s a $j \in J$ with $i \le f(j)$). A net is eventually in $S$ if for some $i \in I$, any upper bound of $i$ is in $S$. A net converges to a point $x$, and we write $x_i \to x$ if it is eventually in $V$ for every neighborhood $V$ of $x$. We call $x$ the limit of the net. A net is cofinally in $S$ if for all $i$, there’s a $j$ with $i \le j$ and $j \in S$. A point $x$ is said to be a cluster point if the net is cofinally in $V$ for all $V$ neighborhoods of $x$.

Proposition 4. $\:$ C $\implies$ Every net has a cluster point $\implies$ Every net has a converging subnet $\implies$ C (uses A.C.)

Proof. (1st $\implies$) Suppose $\{x_i\}_{i \in I}$ has no cluster point. Let $\mathcal{V}(x) = \{ V \text{ neighborhood of } x \ : \ \exists i \in I \forall j \ge i \ \ x_j \notin V \}$. Since no $x$ is a cluster point, $\mathcal{V}(x) \ne \emptyset$. Then $\mathcal{V} = \bigcup_{x \in X} \mathcal{V}(x)$ is an open cover. Take $\{V_n\} \subseteq \mathcal{V}$ finite subcover. For each $n$, there is some $i_n$ for which $j \ge i_n \implies x \notin V_n$. Take any upper bound $j$ of $\{i_n\}$.

(2nd $\implies$) Let $x$ be cluster point of $\{x_i\}_{i \in I}$ and $\mathcal{V}_x$ a basis for $x$. Then $D = \{ (i, V) \in I \times \mathcal{V}_x \ : \ x_i \in V \}$ is a directed set if we define $(i, U) \le (j, V) \iff (i \ge j \text{ and } V \supseteq U)$. Let $\pi : D \to I$ be the projection to $I$. It is order-preserving by definition and cofinal since $x$ is cluster. Finally, it it follows that $x_{f(i, V)} \to x$.

(3rd $\implies$) Let $\{V_i\}_{i \in I}$ be an open cover. Define $F = \{ A \subseteq I \ : \ A \text{ is finite}\}$, this is a directed set. Contrapositively, if it has no finite subcover, take a choice function $x : A \to X$ where $x_A \notin \bigcup_{i \in A} V_i$. The net $\{x_A\}$ cannot have a cluster point: If $V_i$ is a neighborhood of $x$, $x_A \notin V_i$ for any $A \supseteq \{i\}$. $\quad\square$

Definition. A point $p$ is said to be a cluster point of a sequence $(x_n)$ if for any given nbd $V$ of $p$, there are infinitely many $n$ of which $x_n \in V$.

Note that every sequence is itself, a net, as the natural numbers are a directed set. The cluster point notion of nets is the same as above when applied to sequences. Similarly, a subsequence is defined by a function $f : \N \to \N$ which is strictly increasing (as a consequence of being order-preserving and cofinal). A sequence converging to a point is the same notion to convergence of nets when applied to sequences. In a way, sequences are a special case of nets, which has a lot more structure but is also a lot more restricting. In general compact sets, they bring non-equivalent notions.

Lemma 1. If a sequence has a converging subsequence, then it has a cluster point.

Proof. Let $x_{f(n)} \to p$. For any $V$ nbd of $p$, there exists $n_0$ such that $x_{f(n)} \in V$ for $n \ge n_0$, and there are infinitely many such $n$. So $p$ is a cluster point. $\quad\square$

Proposition 5. $\:$ CC $\iff$ Every sequence has a cluster point.

Proof. ($\implies$) Suppose some sequence $(x_n)$ has no cluster point. So every $p$ has a nbd $V$ such that $V \cap \{x_n\}$ is finite. Let $\mathcal{V}_n$ denote all open sets $V$ such that $V \cap \{x_n\} \subseteq \{x_1, \dots, x_n\}$. If $V_n = \bigcup \mathcal{V}_n$, then $\{V_n\}$ is a countable cover of $X$. It can’t have a finite cover, since $x_{n+1} \notin V_n$. ($\impliedby$) Suppose some countable cover $\{V_n\}$ has no finite subcover. Then choose $x_1 \in V_1$ and $x_{n+1} \in X \setminus \bigcup_{k=1}^n V_k$. No $p$ could be a cluster point of $(x_n)$, since $p \in V_n$ for some $n$ and $x_m \notin V_n$ for any $m > n$. $\quad\square$

Definition. Given $A \subseteq X$, its sequential closure $\text{scl}(A)$ is the set of all $p \in X$ of which there exists some sequence $(x_n)$ in $A$ such that $x_n \to p$. A set is sequentially closed if $\text{scl}(A) \subseteq A$.

Lemma 2. $\text{scl}(A) \subseteq \cl{A}$ for any set $A$. Every closed set is sequentially closed.

Proof. If $(x_n) \subseteq A$ with $x_n \to p$, then by definition, for any nbd $V$ of $p$ there is an $n_0$ such that $x_n \in V$ for $n \ge n_0$. So $V \cap \{x_n\} \ne \emptyset$ and therefore, $p \in \cl{A}$. The second fact follows from the fact that a set is closed if and only if $\cl{A} \subseteq A$. $\quad\square$

Definition. A space is sequential (Seq) if every sequentially closed is closed. A space is sequentially compact (SC) if every sequence has a converging subsequence.

Proposition 6. $\:$ SC $\implies$ CC

Proof. Immediate from Lemma 1 and Proposition 5. $\quad\square$

This proves being countably compact is weaker than being compact or sequentially compact. However, both of the latter are incompatible, in the sense that there are topological spaces which are compact and not sequentially compact (using the axiom of choice, consider the product topology on $[0,1]^{[0,1]}$), and sequentially compact spaces which are not compact (The order topology on $\omega_1$, the first uncountable ordinal). Conversely, there’s something beautiful about being able to say that a space is sequentially compact if it is sequential and compact. This is true, but the condition is not necessary, we only require countable compactness.

Proposition 7. $\:$ (CC and Seq) $\implies$ SC

Proof. See Theorem 223 here. $\quad\square$