Compact Theorem 1

 

Thm 1

Let $(X, \mathscr{T})$, $(Y, \mathscr{T}')$ be topological space  and  $A \subseteq X$.

Let $f : X \to Y$ be continuous  and  $A$ be compact. Then

$f(A)$ is compact

 

  Let $\left\{U_{\alpha} \in \mathscr{T}' \; | \; \alpha \in I \right\}$ be open cover of $f(A)$. So

  $f(A) \subseteq \displaystyle \bigcup_{\alpha \in I} U_{\alpha} $

 

  Since $f$ is continuous,

  $f^{-1}(U_{\alpha}) \in \mathscr{T}$

  Also

  $A \subseteq f^{-1}(f(A)) \subseteq f^{-1}(\displaystyle \bigcup_{\alpha \in I} U_{\alpha}) = \displaystyle \bigcup_{\alpha \in I} f^{-1}(U_{\alpha})$

  $\therefore$  $\left\{f^{-1}(U_{\alpha}) \; | \; \alpha \in I \right\} $ is open cover of $A$

 

 

  Since $A$ is compact, there exists $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$  s.t.

  $A \subseteq \displaystyle \bigcup_{i=1}^{n} f^{-1}(U_{\alpha_{i}})$

 

  Since $f(f^{-1}(U_{\alpha_{i}})) \subseteq U_{\alpha_{i}}$,

  $f(A) \subseteq \displaystyle f(\bigcup_{i=1}^{n} f^{-1}(U_{\alpha_{i}})) = \bigcup_{i=1}^{n} f(f^{-1}(U_{\alpha_{i}})) \subseteq \bigcup_{i=1}^{n} U_{\alpha_{i}}$

 

  $\therefore$  $f(A)$ is compact

 

 

 

Cor 1

Let $X$, $Y$ be topological space.

Let $f : X \to Y$ be onto continuous  and  $X$ be compact. Then

$Y$ is compact

 

 

 

Cor 2

Let $X$, $Y$ be topological space  and  $X \cong Y$. Then

$X$ is compact  $\Leftrightarrow$  $Y$ is compact

 

 

 

Thm 2

Let $(X, \mathscr{T})$ be a compact space  and  $A \subseteq X$. Then

$A$ is closed set in $X$  $\Rightarrow$  $A$ is compact

 

  Let $\left\{U_{\alpha} \in \mathscr{T} \; | \; \alpha \in I \right\}$ be an open cover of $A$. So

  $A \subseteq \displaystyle \bigcup_{\alpha \in I} U_{\alpha}$

 

  Consider $U \cup \left\{X \setminus A \right\}$

  Since $A$ is closed set in $X$,

  $X \setminus A \in \mathscr{T}$

  Also

  $(\displaystyle \bigcup_{\alpha \in I} U_{\alpha}) \cup (X \setminus A) = X$

  $\therefore$  $\left\{U_{\alpha} \; | \; \alpha \in I \right\} \cup \left\{X \setminus A \right\}$ is open cover of $X$

 

 

  Since $X$ is compact, there exists $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$  s.t.

  $X = \displaystyle (\bigcup_{i=1}^{n} U_{\alpha_{i}}) \cup (X \setminus A)$     or     $X = \displaystyle (\bigcup_{i=1}^{n} U_{\alpha_{i}}) $

 

  Therefore

  $A \subseteq \displaystyle \bigcup_{i=1}^{n} U_{\alpha_{i}}$

 

  $\therefore$  $A$ is compact

 

 

 

Thm 3

Let $X$ be a Hausdorff topological space  and  $A \subseteq X$. Then

$A$ is compact  $\Rightarrow$  $A$ is closed set in $X$

 

  $\forall p \in X \setminus A$

 

  For each $x \in A$,

  since $X$ is Hausdorff, there exists $U_{x}, V_{x} \in \mathscr{T}$  s.t.

  $p \in U_{x}$,     $x \in V_{x}$,     $U_{x} \cap V_{x} = \varnothing$

  $\therefore$  $\left\{V_{x} \; | \; x \in A \right\}$ is an open cover of $A$.

 

 

  Since $A$ is compact, there exists $x_{1}, x_{2}, \cdots, x_{n} \in A$  s.t.

  $A \subseteq \displaystyle \bigcup_{i=1}^{n} V_{x_{i}}$

 

  Since $V_{x_{i}} \cap U_{x_{i}} = \varnothing$,

  $(\displaystyle \bigcup_{i=1}^{n} V_{x_{i}}) \cap (\bigcap_{i=1}^{n} U_{x_{i}}) = \varnothing$

  $\therefore$  $p \in \displaystyle \bigcap_{i=1}^{n} U_{x_{i}} \subseteq X \setminus A$

 

  Since $\displaystyle \bigcap_{i=1}^{n} U_{x_{i}}$ is open set in $X$, by Prop 2

  $X \setminus A$ is open set in $X$

 

  $\therefore$  $A$ is closed set in $X$

 

 

 

Thm 4

Let $X$ be compact  and  $Y$ be a Hausdorff.

Let $f : X \to Y$ be 1-1 onto continuous. Then

$f$ is homeomorphism

 

  Show : $f^{-1}$ is continuous

  Let $C$ be a closed set in $X$

 

  Since $X$ is compact, by Thm 2

  $C$ is compact

  Since $f$ is continuous, by Thm 1

  $(f^{-1})^{-1}(C) = f(C) $ is compact

  Since $Y$ is Hausdorff, by Thm 3

  $f(C)$ is closed set in $Y$

 

  By Thm 2 - 5,

$f^{-1}$ is continuous

 

  end

 

  $\therefore$  $f$ is homeomorphism

 

 

 

Thm 5

Let $X$, $Y$ be a topological space.

T.F.A.E.

$(1)$  $X \times Y$ is compact

$(2)$  $X$  and  $Y$ are compact

 

pf)
$(1) \Rightarrow (2)$

  Consider projective map $P_{1} : X \times Y \to X$,  $P_{2} : X \times Y \to Y$

  By Prop 1,

  $P_{1}$  and  $P_{2}$ are continuous

 

  By assumption and Thm 1,

  $P_{1}(X \times Y) = X$  and  $P_{2}(X \times Y) = Y$ are compact

 

$(2) \Leftarrow (1)$

  Let $D = \left\{U_{\alpha} \times V_{\alpha} \; | \; \alpha \in I \right\}$ be open cover of $X \times Y$. So

  $X \times Y = \displaystyle \bigcup_{\alpha \in I} (U_{\alpha} \times V_{\alpha})$

 

  For each $x \in X$,

  since $Y \cong \left\{x \right\} \times Y$ and $Y$ is compact, by Cor 2

  $\left\{x \right\} \times Y$ is compact

 

  there exists $\alpha(x, 1), \alpha(x, 2), \cdots, \alpha(x, n_{x}) \in I$  s.t.

  $\left\{x \right\} \times Y \subseteq \displaystyle \bigcup_{j=1}^{n_{x}} ( U_{\alpha(x, j)} \times V_{\alpha(x, j)} )$

  &     $x \in U_{\alpha(x, j)}$

 

  Let $T_{x} = \displaystyle \bigcap_{j=1}^{n_{x}} U_{\alpha(x, j)}$

  So $T_{x}$ is open set in $X$  and  $x \in T_{x}$

  $\therefore$  $T = \left\{T_{x} \; | \; x \in X \right\}$ is open cover of $X$

 

 

  Since $X$ is compact, there exists $x_{1}, x_{2}, \cdots, x_{m} \in X$  s.t.

  $X = \displaystyle \bigcup_{i=1}^{m} T_{x_{i}}$

  Therefore

  $X \times Y = \displaystyle (\bigcup_{i=1}^{m} T_{x_{i}}) \times Y = \bigcup_{i=1}^{m} (T_{x_{i}} \times Y)$

 

  Since $Y = \displaystyle \bigcup_{j = 1}^{n_{x_{i}}} V_{\alpha(x_{i}, j)}$,

  $T_{x_{i}} \times Y = \displaystyle \bigcap_{j=1}^{n_{x_{i}}} U_{\alpha(x_{i}, j)} \times \bigcup_{j=1}^{n_{x_{i}}} V_{\alpha(x_{i}, j)}$

                   $= \displaystyle \bigcup_{j=1}^{n_{x_{i}}} (\bigcap_{j=1}^{n_{x_{i}}} U_{\alpha(x_{i}, j)} \times  V_{\alpha(x_{i}, j)})$

                   $\subseteq \displaystyle \bigcup_{j=1}^{n_{x_{i}}} (U_{\alpha(x_{i}, j)} \times V_{\alpha(x_{i}, j)})$ 

 

 

  Therefore

  $X \times Y = \displaystyle \bigcup_{i=1}^{m} (T_{x_{i}} \times Y)
\subseteq \bigcup_{i=1}^{m} \bigcup_{j=1}^{n_{x_{i}}} (U_{\alpha(x_{i}, j)} \times V_{\alpha(x_{i}, j)}) \subseteq X \times Y$

 

  $\therefore$  $\left\{U_{\alpha(x_{i}, j)} \times V_{\alpha(x_{i}, j)} \; | \; 1 \leq i \leq m, \; 1 \leq j \leq n_{x_{i}} \right\}$ is a finite subcover of $D$

 

  By Thm 1,

  $X \times Y$ is compact

 

 

 

Cor 3

Let $X_{i}$ be compact  for $i = 1, 2, \cdots, n$. Then

$X_{1} \times \cdots \times X_{n}$ is compact

 

 

 

Thm 6

Let $(X, \mathscr{T})$ be a topological space  and  $A, B \subseteq X$ are compact. Then

$A \cup B$ is compact

 

  Let $\left\{U_{\alpha} \in \mathscr{T} \; | \; \alpha \in I \right\}$ be open cover of $A \cup B$. So

  $A \cup B \subseteq \displaystyle \bigcup_{\alpha \in I} U_{\alpha}$

  $\therefore$  $\left\{U_{\alpha} \; | \; \alpha \in I \right\}$ be open cover of $A$, $B$

 

 

  Since $A$ is compact, there exists $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \in I$  s.t.

  $A \subseteq \displaystyle \bigcup_{i=1}^{n} U_{\alpha_{i}}$

  Since $B$ is compact, there exists $\alpha_{n+1}, \alpha_{n+2}, \cdots, \alpha_{n+m} \in I$  s.t.

  $B \subseteq \displaystyle \bigcup_{i = n+1}^{n + m} U_{\alpha_{i}}$

 

  Therefore

  $A \cup B \subseteq \displaystyle \bigcup_{i = 1}^{n + m} U_{\alpha_{i}}$

 

  $\therefore$  $A \cup B$ is compact

 

 

 

Cor 4

Let $(X, \mathscr{T})$ be a topological space  and  $A_{i} \subseteq X$ are compact  for $i = 1, 2, \cdots, n$. Then

$\displaystyle \bigcup_{i=1}^{n} A_{i}$ is compact