İki topolojik uzaydan yeni bir topolojik uzay oluşturmak-II

Question

$(X,\tau_1),(Y,\tau_2)$ topolojik uzaylar ve $Z=X\cup Y$ olmak üzere

$$\tau_3:=\{A\subseteq Z|(A\cap X\in\tau_1)(A\cap Y\in\tau_2)\}\subseteq 2^Z$$ ailesinin $Z$ kümesi üzerinde bir topoloji olduğunu gösteriniz.

Answer 1

$\mathbf{T_1)}$ $\emptyset,Z\overset{?}{\in}\tau_3$

$\left.\begin{array}{rr}\emptyset\cap X=\emptyset \in\tau_1\\ \\ \emptyset\cap Y=\emptyset \in\tau_2 \end{array}\right\}\Rightarrow \emptyset\in\tau_3$                 $\left.\begin{array}{rr} Z\cap X =X \in\tau_1\\ \\ Z\cap Y=Y \in\tau_2 \end{array}\right\}\Rightarrow Z\in\tau_3$

 

$\mathbf{T_2)}$ $A,B\in \tau_3$ olsun.

$\left.\begin{array}{rr} A\in \tau_3\Rightarrow (A\cap X\in \tau_1)(A\cap Y\in \tau_2) \\  \\ B\in \tau_3\Rightarrow (B\cap X\in \tau_1)(B\cap Y\in \tau_2) \end{array}\right\}\Rightarrow $

$\Rightarrow ((A\cap X)\cap (B\cap X)\in \tau_1)((A\cap Y)\cap (B\cap Y)\in \tau_2)$
 

$\Rightarrow ((A\cap B)\cap X\in \tau_1)((A\cap B)\cap Y\in \tau_2)$

 
$\Rightarrow A\cap B\in\tau_3.$

$\mathbf{T_3)}$ $\mathcal{A}\subseteq \tau_3$ olsun.

$\begin{array}{rcl}\mathcal{A}\subseteq \tau_3 & \Rightarrow & (\forall A\in\mathcal{A})(A\cap X\in\tau_1)(A\cap Y\in\tau_2) \\  \\ & \Rightarrow & \left(\bigcup_{A\in\mathcal{A}}(A\cap X)\in\tau_1\right)\left(\bigcup_{A\in\mathcal{A}}(A\cap Y)\in\tau_2\right) \\ \\ & \Rightarrow & ((\bigcup\mathcal{A})\cap X\in\tau_1)((\bigcup\mathcal{A})\cap Y\in\tau_2) \\  \\ & \Rightarrow & \bigcup\mathcal{A}\in\tau_3\end{array}$