# Random math: 33

Source of most content: Gunhee Cho.

# Product of Compact Spaces

$\text{Let } X_i \text{ be compact space for } i=1,\ldots, n. \text{ Then, } \prod_{i=1}^n X_i \text{ is compact.} \nonumber$

## Tube Lemma

$\text{Suppose } X, Y \text{ be compact. For any } x_0 \in X \text{ and nbhd } U \text{ of } \{x_0 \} \times Y \, \exists \text{ nbhd } W \text{ of } x_0 \in X \text{ s.t. } W \times Y \subset U. \nonumber$

### Proof (Tube)

Let $x_0$ be fixed. For each $(x_0, y) \in X \times Y$, choose $U_y \times V_y$, a nbhd of $(x_0, y)$ where $U_y \in T_X$ and $V_y \in T_Y$. Then, $\{U_x \times V_y \subset U: y \in Y \}$ is an open cover of $\{x_0\} \times Y$. If we consider the projection map $p: X\times Y \to Y$ which is onto and continous, $\{ x_0 \} \times Y$ is homeomorphic to $Y$ since $p^{-1}$ is also continous for the fixed $x_0$. Therefore, $\{x_0\}\times Y$ is compact since $Y$ is compact by assumption. Then, we can choose $y_1, \ldots, y_n \in Y$ s.t. $\{x_0 \} \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k})$. Set $W := \cap_{k=1}^n U_{y_k}$ that contains $x_0$ and is open. Then, $\{ x_0 \} \times Y \subset W \times Y \subset \cup_{k=1}^n (U_{y_k} \times V_{y_k}) \subset U$.

### Proof

It suffices to show that for compact sets $X,Y$, $X\times Y$ is compact. Let $\{ A_{\alpha^{x}} : \alpha \in J \}$ be an open cover of $\{x \} \times Y$ for the fixed $x \in X$. Since $\{ x \} \times Y$ is compact, there exists finite open subcover s.t. $\{x \}\times Y \subset \cup_{k=1}^n A_{\alpha_k}$. By Tube lemma, there is $W_x$ s.t. $W_x \times Y \subset \cup_{k=1}^n A_{\alpha_k^x}$. Then, $\{ W_{x} : x \in X\}$ will be an open cover of $X$. Since $X$ is compact, $\exists x_1, \ldots, x_m \in X$ s.t. $X \subset \cup_{k=1}^m W_{x_k}$. Therefore, $X\times Y \subset (\cup_{k=1}^m W_{x_k}) \times Y = \subset \cup_{k=1}^m (W_{x_k} \times Y) \stackrel{Tube}{\subset} \cup_{k=1}^m (\cup_{l=1}^n A_{\alpha_{l}^{x_k}})$. Hence, $X\times Y$ is covered by at most finite open sets.

NB1; In fact, the infinite product of compact spaces is still compact.(Tychonoff’s theorem).
NB2; Thus $[ 0, 1]^2$ is compact and $[0, 1]^{[0, 1 ]}$ is compact.

# Finite Intersection Property

Def. Let $X$ be a set and $C \subset 2^X$. Then, $C$ has the finite intersection property (FIP) if for any $c_1, \ldots, c_n \in C$, $\cap_{k=1}^n C_n \ne \emptyset$.

## Proposition

$X \text{ is compact iff any } C \subset 2^X \text{ which has 1. FIP 2. for any } c \in C, c \text{ is closed in }X \text{ satisfies } \cup_{c \in C} c \ne \emptyset. \nonumber$

### Proof

1. $(\Rightarrow)$ If not, then $\exists C$ s.t. $C$ satisfies 1 and 2 but $\cup_{c \in C} c = \emptyset$, which implies $\cup_{c \in C} c^c = X$. Since $X$ is compact and $c^c$ is open, $\bigcup_{finite \, c} c^c = X$ holds. Then, $\bigcap_{finite \, c} c = \emptyset$ holds, i.e., it contradicts to FIP, which concludes the proof.
2. $(\Leftarrow)$ Suppose not. Then, $\exists$ open cover $\{ O _\alpha \} _{\alpha \in J}$ of $X$ s.t. $X = \cup _{\alpha \in J} O _\alpha$ but $\cup _{finite \, \alpha} O _\alpha \subsetneq X$ for any choice of finite indices. Then, $\cap _{finite \, \alpha} O _\alpha^c \ne \emptyset$. Since $O _\alpha^c$ is closed, $\{ O _\alpha^c : \alpha \in J\}$ has FIP. By Asm, $\cup _{\alpha \in J} O _\alpha^c \ne \emptyset$, i.e., $\cup _{\alpha \in J} O _\alpha \ne X$ which is a contradiction.