Hausdorff Uzaylarının Karakterizasyonuna Dair-III

Question

Bir $(X,\tau)$ topolojik uzayının Hausdorff uzayı olması için gerek ve yeter koşul uzayın her $x$ noktası için $x$ noktasının kapalı komşuluklarının arakesitinin sadece $x$ noktasından ibaret olmasıdır. Biçimsel olarak

$$((X,\tau),  \text{ topolojik uzay})(\mathcal{K}(x):=\{K|(K\in\mathcal{N}(x))(K\in\mathcal{C}(X,\tau))\})$$

$$\Rightarrow$$

$$(X,\tau), \text{ Hausdorff uzayı}\Leftrightarrow (\forall x\in X)(\{x\}=\cap\mathcal{K}(x))$$

şeklinde ifade edilir.

Answer 1

$\left( \Rightarrow \right) :\left( X,\tau \right) , \ T_{2}$ uzayı ve $x\in X$ olsun.

$\left. \begin{array}{rr} x\in X \\ \\ \mathcal{K}\left( x\right):=\left\{ K|\left( K\in \mathcal{N}\left( x\right) \right) \left( K\in \mathcal{C}\left( X,\tau \right) \right) \right\} \end{array} \right\} \Rightarrow \left\{ x\right\} \subseteq \cap \mathcal{K}\left( x\right) \ldots \left( 1\right) $ 

Şimdi de $y\in \cap \mathcal{K}\left( x\right)$ iken $y\in \left\{ x\right\} $ veya bununla aynı anlama gelen $y\notin \left\{ x\right\} $ iken $y\notin \cap \mathcal{K}\left( x\right) $ olduğunu göstermeliyiz. $\left. \begin{array}{r} \left( x\in X\right) \left( y\notin \left\{ x\right\} \right) \Rightarrow \left( x,y\in X\right) \left( x\neq y\right) \\ \\ \left( X,\tau \right) ,\text{ }T_{2}\text{ uzayı } \end{array} \right\} \Rightarrow \begin{array}{c} \mbox{} \\ \mbox{} \\ \left. \begin{array}{r} \left( \exists U\in \mathcal{U}\left( x\right) \right) \left( y\notin \overline{U}\right) \\ \\ K:=\overline{U} \end{array} \right\} \Rightarrow \end{array}$

$\left. \begin{array}{c} \end{array}\right.$ $\left. \begin{array}{c} \Rightarrow \left( K\in \mathcal{N}\left( x\right) \right) \left( K\in \mathcal{C}\left( X,\tau \right) \right) \left( y\notin K\right) \end{array}\right.$

$\left. \begin{array}{c} \Rightarrow y\notin \cap \mathcal{K}\left( x\right) \end{array}\right. $ 

$\left. \begin{array}{c} \text{O halde }\cap \mathcal{K}\left( x\right) \subseteq \left\{ x\right\} \ldots \left( 2\right) \end{array}\right. $ 

$\left. \begin{array}{c} \left( 1\right) ,\left( 2\right) \Rightarrow \left\{ x\right\} =\cap \mathcal{K}\left( x\right). \end{array}\right.$

$\left( \Leftarrow \right):x,y\in X$ ve $x\neq y$ olsun.

$\left. \begin{array}{r} \left( x,y\in X\right) \left( x\neq y\right) \\ \\ \text{Hipotez} \end{array} \right\} \Rightarrow y\notin \left\{ x\right\} =\cap \mathcal{K}\left( x\right) \in \mathcal{C}\left( X,\tau \right) $

$\left. \begin{array}{r} \Rightarrow \left( \exists K\in \mathcal{K}\left( x\right) \right) \left( y\notin K\right) \left( y\in \left( X\setminus K\right) \in \tau \right) \\ \\ V:=X\setminus K \end{array} \right\} \Rightarrow $

$\left. \begin{array}{c} \Rightarrow \left( y\notin K\in \mathcal{N}\left( x\right) \right) \left( V\in \mathcal{U}\left( y\right) \right) \end{array} \right. $

$\left. \begin{array}{c} \Rightarrow \left( \exists U\in \mathcal{U}\left( x\right) \right) \left( y\notin K\supseteq U\right) \left( V\in \mathcal{U}\left( y\right) \right) \end{array} \right. $

$\left. \begin{array}{c} \Rightarrow \left( U\in \mathcal{U}\left( x\right) \right) \left( V\in \mathcal{U}\left( y\right) \right) \left( V=X\setminus K\subseteq X\setminus U\right) \end{array} \right. $

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