Question

$(X,\tau)$ topolojik uzay ve $Y\subseteq X$ olsun. $$\overline{Y}^{\circ}=\emptyset\Leftrightarrow (\forall U\in \tau\setminus\{\emptyset\})(\exists V\in\tau\setminus\{\emptyset\})(V\subseteq U)(V\cap Y=\emptyset).$$

Answer 1

$(\Rightarrow):$ $\overline{Y}^{\circ}=\emptyset$ olsun.

$\overline{Y}^{\circ}=\emptyset \Rightarrow \bigcup\{U|(U\subseteq \overline{Y})(U\in\tau)\}=\emptyset$

$\Rightarrow \{U|(U\subseteq \overline{Y})(U\in\tau)\}=\{\emptyset\}$

$\Rightarrow (\forall U\in\tau\setminus \{\emptyset\})(U\nsubseteq\overline{Y})$

$\Rightarrow (\forall U\in\tau\setminus \{\emptyset\})(\exists x\in U)(x\notin\overline{Y})$

$\left.\begin{array}{rcl}\Rightarrow (\forall U\in\tau\setminus \{\emptyset\})(\exists x\in U)(\exists W\in\mathcal{U}(x))(W\cap Y=\emptyset) \\ \\ V:=W\cap U\end{array}\right\}\Rightarrow$

$\Rightarrow (\forall U\in\tau\setminus \{\emptyset\})(\exists V\in \tau\setminus\{\emptyset\})(V\subseteq U)(V\cap Y=\emptyset).$

 

$(\Leftarrow):$ Amacımız  $\overline{Y}^{\circ}=\emptyset$  olduğunu göstermek. Bunun için $$(\forall x\in X)\left(x\notin \overline{Y}^{\circ}\right)$$ önermesinin doğru olduğunu yani $$(\forall x\in X)(U\in\mathcal{U}(x)\Rightarrow U\nsubseteq \overline{Y})$$ önermesinin doğru olduğunu göstermeliyiz.

$x\in X$  ve  $U\in\mathcal{U}(x)$  olsun. $U\nsubseteq \overline{Y}$ olduğunu gösterirsek kanıt biter.

$\left.\begin{array}{rr} (x\in X)(U\in\mathcal{U}(x))\Rightarrow U\in\tau\setminus\{\emptyset\} \\ \\ \text{Hipotez}\end{array}\right\}\Rightarrow$

$\Rightarrow (\exists V\in\tau\setminus\{\emptyset\})(V\subseteq U)(V\cap Y=\emptyset)$

$\Rightarrow (\exists V\in\tau\setminus\{\emptyset\})(V\subseteq U)(V\cap \overline{Y}\subseteq \overline{V\cap Y}=\overline{\emptyset}=\emptyset)$

$\Rightarrow (\exists V\in\tau\setminus\{\emptyset\})(V\subseteq U)(V\cap \overline{Y}=\emptyset)$

$\Rightarrow (V\subseteq U)(V\nsubseteq \overline{Y})$

$\Rightarrow U\nsubseteq \overline{Y}.$