Compact Theorem 2
Thm 1 Bolzano - Weierstrass Theorem
Let $(X, \mathscr{T})$ be a compact space and $A$ be an infinite subset in $X$. Then
$A$ has a limit point in $X$.
Suppose : $A$ don't have a limit point in $X$
So for each $x \in X$, there exists $U_{x} \in \mathscr{T}$ containing $x$ s.t.
$(U_{x} \setminus \left\{x \right\}) \cap A = \varnothing$
$\therefore$ $\left\{U_{x} \; | \; x \in X \right\}$ is open cover of $X$.
Since $X$ is compact, there exists $x_{1}, x_{2}, \cdots, x_{n} \in X$ s.t.
$X = \displaystyle \bigcup_{i=1}^{n} U_{x_{i}}$
Therefore
$A = X \cap A = (\displaystyle \bigcup_{i=1}^{n} U_{x_{i}} ) \cap A= \bigcup_{i=1}^{n} (U_{x_{i}} \cap A)$
Since $(U_{x_{i}} \setminus \left\{x_{i} \right\}) \cap A = \varnothing$,
$A$ is finite set
Contradiction
Thm 2
Let $A \subseteq (\mathbb{R}^{n}, \text{usual})$
T.F.A.E.
$(1)$ $A$ is compact
$(2)$ $A$ is closed set in $\mathbb{R}^{n}$ and bounded
pf)
$(1) \Rightarrow (2)$
Since $\mathbb{R}^{n}$ is Hausdorff and $A$ is compact, by Thm 3
$A$ is closed set in $\mathbb{R}^{n}$
Consider $\left\{B_{n}(O) \; | \; n \in \mathbb{N} \right\}$, the open cover of $A$
Since $A$ is compact, there exists $N \in \mathbb{N}$ s.t.
$A \subseteq B_{N}(O)$
$\therefore$ $A$ is bounded
$(2) \Leftarrow (1)$
Since $A$ is bounded, there exists $s > 0$ s.t.
$A \subseteq [-s, s]^{n}$
$\therefore$ $A = A \cap [-s, s]^{n}$
Since $A$ is closed set in $\mathbb{R}^{n}$, by Thm 2 - 3
$A$ is closed set in $[-s, s]^{n}$
By Cor 3,
$[-s, s]^{n}$ is compact
Since $A$ is closed set in $[-s, s]^{n}$, by Thm 2
$A$ is compact
Thm 3
Let $X$ be a compact space and $f : X \to \mathbb{R}$ be continuous. Then
$f$ achieves maximum value and minimum value.
Since $f$ is continuous and $X$ is compact, by Thm 1
$f(X)$ is compact
By Thm 2,
$f(X)$ is closed set in $\mathbb{R}$ and bounded
$\therefore$ $f(X)$ has a least upper bound $M$ and greatest lower bound $m$.
Since $f(X)$ is closed set,
$m, M \in \overline{f(X)} = f(X)$
So there exists $x, y \in X$ s.t.
$f(x) = m$, $f(y) = M$
$\therefore$ $f$ achieves maximum value and minimum value.
Def 1
Let $X$ be a topological space.
Let $\mathscr{C} = \left\{C_{\alpha} \; | \; \alpha \in I \right\}$ be a collection of closed set in $X$.
$\mathscr{C}$ satisfies finite intersection property if
$\forall \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$, $\displaystyle \bigcap_{i = 1}^{n} C_{\alpha_{i}} \neq \varnothing$
Thm 4
Let $(X, \mathscr{T})$ be a topological space.
T.F.A.E.
$(1)$ $X$ is compact
$(2)$ Let $\mathscr{C}$ be a collection of closed set in $X$.
$\mathscr{C}$ satisfies finite intersection property $\Rightarrow$ $\displaystyle \bigcap_{C \in \mathscr{C}} C \neq \varnothing$
pf)
$(1) \Rightarrow (2)$
Let $\mathscr{C}$ be a collection of closed set in $X$ satisfying finite intersection property.
Suppose : $\displaystyle \bigcap_{C \in \mathscr{C}} C = \varnothing$
So $\displaystyle \bigcup_{C \in \mathscr{C}} (X \setminus C) = X \setminus (\bigcap_{C \in \mathscr{C}} C) = X$
$\therefore$ $\left\{X \setminus C \; | \; C \in \mathscr{C} \right\}$ is open cover of $X$
Since $X$ is compact, there exists $C_{1}, C_{2}, \cdots, C_{n} \in \mathscr{C}$ s.t.
$X = \displaystyle \bigcup_{i=1}^{n} (X \setminus C_{i})$
$\therefore$ $\displaystyle \bigcap_{i=1}^{n} C_{i} = \varnothing$
Contradiction
$(2) \Leftarrow (1)$
Suppose : $X$ is not compact
So there exists $U = \left\{U_{\alpha} \; | \; \alpha \in I \right\}$ be open cover of $X$ s.t.
$U$ don't have finite subcover
So
$\forall \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$, $\displaystyle \bigcup_{i=1}^{n} U_{\alpha_{i}} \neq X$
Let $\mathscr{C} = \left\{X \setminus U_{\alpha} \; | \; \alpha \in I \right\}$
So $\mathscr{C}$ be collection of closed set in $X$.
Also $\forall \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$,
$\displaystyle \bigcap_{i = 1}^{n} (X \setminus U_{\alpha_{i}}) = X \setminus (\bigcup_{i = 1}^{n} U_{\alpha_{i}}) \neq \varnothing$
$\therefore$ $\mathscr{C}$ satisfies finite intersection property.
$\therefore$ $\displaystyle \bigcap_{\alpha \in I} (X \setminus U_{\alpha}) \neq \varnothing$
$\therefore$ $\displaystyle \bigcup_{\alpha \in I} U_{\alpha} \neq X$
Contradiction
Def 2
Let $(X, \mathscr{T})$ be topological space.
Let sequence $\left\{x_{n} \right\}$ in $X$ by
$x_{n} : \mathbb{N} \to X$
We define
$\displaystyle \lim_{n \to \infty} x_{n} = x$
$\Leftrightarrow$ $\forall V \in \mathscr{T}$ containing $x$, $\exists N >0$ s.t. $n \geq N$ $\Rightarrow$ $x_{n} \in V$
So
$\displaystyle \lim_{n \to \infty} x_{n} \neq x$
$\Leftrightarrow$ $\exists V \in \mathscr{T}$ containing $x$, $\forall N > 0$, $\exists M \geq N$ s.t. $x_{M} \notin V$
Def 3
Let $X$ be a topological space.
1. $X$ has the Bolzano-Weierstress property if
every infinite subset of $X$ has a limit point in $X$.
$X$ : limit point compact space
2. $X$ is a sequentially compact space if
every sequence has a convergent subsequence in $X$.