# Random math: 26

Source of most content: Gunhee Cho and Lecture notes by Parimal Parag.

# Cover and Compact

Def. Given a T.S. $(X, T_X)$, $\{ A_\alpha \}_{\alpha \in J} \subset 2^X$ is a cover of $X$ if $X \subset \bigcup _{\alpha \in J} A _\alpha$.
If $\{ A _\alpha \} _{\alpha \in J} \subset T_X$, then it is called open cover of $X$.

Def. A T.S. $(X, T_X)$ is compact if for any open cover of $X$, $\{ U _\alpha \} _{\alpha \in J}$, $\exists \alpha _1, \ldots, \alpha _n$ (finite) s.t. $X \subset \bigcup _{k=1}^n U _{\alpha _k}$, i.e., there exists a finite open subcover of $X$ for any open cover of $X$.

NB1; For finite set $X$, $(X,T)$ is compact.
NB2; $((0, 1), T_{subspace})$ is not compact. $\forall n \in \mathbb{N}$, define $U_n = (\frac{1}{n}, 1)$. Then, $(0,1) = \cup_{n \in \mathbb{N}}(\frac{1}{n}, 1)$, i.e., $\{ U _n \} _{n \in \mathbb{N}}$ is an open cover of $(0,1)$. However, there is no finite subcover s.t. $X \subset \cup _{k=1}^m U _{\alpha_k}$.
NB3; $(\mathbb{R}, T _{std})$ is not compact. Consider $U_n = (n, n+2)$.

## Properties

1. Closed subspace of compact space is compact.
2. Compact subspace of a Hausdorff space is closed.
3. Let $f: X \rightarrow Y$ be cts function. If $X$ is compact, then $f(X)$ is compact.

### Proof

1. Let $(Y, T_Y)$ be a closed subspace of $(X, T_X)$. Since $Y$ is closed, $X-Y \in T_X$. Let $U$ be an open cover of $Y$, which consists of an open sets in $(X, T_X)$. Then, $U \cup \{ X-Y \}$ is an open cover of $X$. Since $(X, T_X)$ is a compact set, there is a finite subcover of $U \cup \{ X-Y \}$, $V = \{ U_1, \ldots, U_m \}$, which covers $X$. If $X-Y \in V$, $V \setminus \{ X-Y \}$ is a finite subcover of $Y$. If not, $V$ is a finite subcover of $Y$. Thus, $(Y, T_Y)$ is compact.
2. Let $X$ be a Hausdorff space and compact subspace $C \subseteq X$. Take $x \in X - C$, then for each $y \in C$, $\exists$ nbhd $U_{x,y}, V_y$ of $x$ and $y$ s.t. $U_{x,y} \cap V_y \stackrel{Hausdorff}{=} \emptyset$. Then, $\{ V_y: y \in C\}$ is an open cover of $C$. Since $C$ is compact, $\exists y_1, \ldots, y_n$ s.t. $C \subset V = \cup_{k=1}^n V_{y_k}$. Take intersection of disjoint nbhd of $x$, i.e., $U_x = \cap_{k=1}^n U_{x,y_k}$. Then, $U_x \cap V = \emptyset$. Thus, $X \setminus C = \cup_{x\in X \setminus C} U_x$ is open, i.e., $C$ is closed.
3. Let $\mathcal{V}$ be an open cover of $f(X)$. Since $f$ is cts, $\forall V \in \mathcal{V}, f^{-1}(V)$ is open in $X$. Then, $\{ f^{-1}(V) : V \in \mathcal{V} \}$ is an open cover of $X$. As $X$ is compact, it has a finite subcover $X \subset \cup_{i=1}^n f^{-1}(V_i)$. Thus, $\{V_1, \ldots, V_n \}$ is a finite subcover of $f(X)$.

$\text{Let } f: X \rightarrow Y \text{ be cts, bijectve fn. Let } X \text{ be compact and } Y \text{ be Hausdorff. Then, } f \text{ is a homeomorphism.} \nonumber$

### Proof

It suffices to show $f^{-1}$ is cts fn. Take any closed $F$ in $X$. Then, by above 1, $F$ is compact. By 3, $f(F)$ is compact. Then, by 2, $f(F)$ is closed.