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
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