Kompakt ve Hausdorff olan her topolojik uzayın bir normal uzay olduğunu gösteriniz.

Question

$(X,\tau)$ topolojik uzay olmak üzere

$$((X,\tau), \text{ kompakt uzay})((X,\tau), \text{ Hausdorff})\Rightarrow (X,\tau), \text{ normal}$$ olduğunu gösteriniz.

Answer 1

$(X,\tau ),$ kompakt uzay; $\left( X,\tau \right) ,$ Hausdorff; $A\in \mathcal{C}\left( X,\tau \right) ;$ $B\in \mathcal{C}\left( X,\tau \right) ,$ $A\cap B=\emptyset,$ $x\in A$ ve $y\in B$ olsun.

$\left.\begin{array}{r}\left( x\in A\right) \left( y\in B\right)  \\ \\ A\cap B=\emptyset \end{array}\right\}\Rightarrow \begin{array}{c}\mbox{} \\ \mbox{} \\ \left.\begin{array}{r} (x,y\in X)(x\neq y) \\ \mbox{} \\ {\left( X,\tau \right) ,\text{ }T_{2}}\end{array}\right\}\Rightarrow \!\!\!\!\!\end{array}$

$\mbox{}$

$\left.\begin{array}{r}\Rightarrow \left( \exists U_{x}\in \mathcal{U}\left( x\right) \right) \left( \exists V_{y}\in \mathcal{U}\left( y\right) \right) \left( U_{x}\cap V_{y}=\emptyset \right) \\ \\ \mathcal{A}:=\left\{ U_{x}|(x,y\in X)(x\neq y)\Rightarrow (\exists U_x\in\mathcal{U}(x))(\exists V_y\in\mathcal{U}(y))(U_x\cap V_y=\emptyset)\right\} \\ \\ \mathcal{B}:=\left\{ V_{y}|(x,y\in X)(x\neq y)\Rightarrow (\exists U_x\in\mathcal{U}(x))(\exists V_y\in\mathcal{U}(y))(U_x\cap V_y=\emptyset)\right\} \end{array}\right\} \Rightarrow$

$\mbox{}$

$\left.\begin{array}{r}\Rightarrow \left( \mathcal{A}\subseteq \tau \right) \left( A\subseteq \cup \mathcal{A}\right)\left( \mathcal{B}\subseteq \tau \right) \left( B\subseteq \cup \mathcal{B}\right)  \\ \\ \left( (X,\tau ),\text{ kompakt uzay}\right) \left( A,B\in \mathcal{C}\left( X,\tau \right) \right) \Rightarrow \left( A,\text{ }\tau \text{-kompakt}\right)\left( B,\text{ }\tau \text{-kompakt}\right)\end{array}\right\} \Rightarrow$

$\mbox{}$

$\left.\begin{array}{r}\Rightarrow \left( \exists \mathcal{A}^{\ast }\subseteq \mathcal{A}\right) \left( \left\vert \mathcal{A}^{\ast} \right\vert <\aleph _{0}\right) \left( A\subseteq \cup \mathcal{A}^{\ast}\right) \left( \exists \mathcal{B}^{\ast }\subseteq \mathcal{B}\right) \left( \left\vert \mathcal{B}^{\ast }\right\vert <\aleph _{0}\right) \left( B\subseteq \cup \mathcal{B}^{\ast}\right)\\ \\ \left( U:=\cup \mathcal{A}^{\ast }\right) (V:=\cup \mathcal{B}^{\ast })\end{array}\right\} \Rightarrow$

$\mbox{}$

$\left.\begin{array}{c}\Rightarrow \left( U\in \mathcal{U}(A)\right)\left( V\in \mathcal{U}(B)\right)\left( U\cap V=\emptyset \right).\end{array}\right.$

 

NOT :  $\mathcal{U}(A):=\{U|(U\in \tau)(A\subseteq U)\}$