Aşağıdaki ailenin bir topoloji olduğunu gösteriniz.

Question

$(X,\tau_1)$ topolojik uzay olmak üzere

$$\tau_2=\left\{A\big{|}(\forall x\in A)(\exists U\in\mathcal{U}(x))(|U\setminus A|\leq\aleph_0)\right\}\subseteq 2^X$$ ailesinin $X$ kümesi üzerinde bir topoloji olduğunu gösteriniz.

Not: $(X,\tau)$ topolojik uzay ve  $x\in X$ olmak üzere $$\mathcal{U}(x):=\{U|x\in U\in \tau\}.$$

Answer 1

$\mathbf{T_1)}$ $\emptyset, X\overset{?}{\in}\tau_2$

$$[(\forall x\in \emptyset)(\exists U\in\mathcal{U}(x))(|U\setminus \emptyset|\leq \aleph_0)]$$$$\equiv$$$$ \forall x[\underset{0}{\underbrace{x\in \emptyset}}\rightarrow \underset{p}{\underbrace{(\exists U\in\mathcal{U}(x))(|U\setminus \emptyset|\leq \aleph_0)}}]\equiv 1$$

olur yani $$(\forall x\in \emptyset)(\exists U\in\mathcal{U}(x))(|U\setminus \emptyset|\leq \aleph_0)$$

önermesi doğru yani $$\emptyset \in\tau_2$$

olur. Öte yandan $x\in X$ olmak üzere

$$\left.\begin{array}{rr} x\in X\\ \\ U:=X \end{array}\right\}\Rightarrow (U\in \mathcal{U}(x))(|U\setminus X|=|X\setminus X|=|\emptyset|=0\leq \aleph_0)$$ olduğundan 

$$(\forall x\in X)(\exists U\in\mathcal{U}(x))(|U\setminus X|\leq \aleph_0)$$ önermesi doğru yani $$X\in \tau_2$$ olur.

$\mathbf{T_2)}$ $A,B\in \tau_2$ ve $x\in A\cap B$ olsun.

$\left.\begin{array}{rr} x\in A\cap B\Rightarrow (x\in A)(x\in B) \\ \\ A,B\in \tau_2\end{array}\right\}\Rightarrow (\exists U\in\mathcal{U}(x))(|U\setminus A|\leq \aleph_0)(\exists V\in\mathcal{U}(x))(|V\setminus B|\leq \aleph_0)$

$\Rightarrow (\exists U\in\mathcal{U}(x))(\exists V\in\mathcal{U}(x))(|U\setminus A|\leq \aleph_0)(|V\setminus B|\leq \aleph_0)$

$\Rightarrow (U\cap V\in\mathcal{U}(x))(|[(U\cap V)\setminus A]\cup [(U\cap V)\setminus B]|\leq |(U\setminus A)\cup (V\setminus B)|\leq \aleph_0)$

$\Rightarrow (U\cap V\in\mathcal{U}(x))(|[(U\cap V)\cap (\setminus A)]\cup [(U\cap V)\cap (\setminus B)]|\leq \aleph_0)$

$\Rightarrow (U\cap V\in\mathcal{U}(x))(|(U\cap V)\cap [(\setminus A)\cup (\setminus B)]|\leq \aleph_0)$

$\Rightarrow (U\cap V\in\mathcal{U}(x))(|(U\cap V)\cap [\setminus (A\cap B)]|\leq \aleph_0)$

$\Rightarrow (U\cap V\in\mathcal{U}(x))(|(U\cap V)\setminus (A\cap B)|\leq \aleph_0).$

$\mathbf{T_3)}$ $\mathcal{A}\subseteq \tau_2$ ve $x\in \cup \mathcal{A}$ olsun.

$$\left.\begin{array}{rr} x\in \cup \mathcal{A}\Rightarrow (\exists A\in\mathcal{A})(x\in A) \\ \\ \mathcal{A}\subseteq \tau_2 \end{array}\right\}\Rightarrow (\exists U\in\mathcal{U}(x))(|U\setminus (\cup \mathcal{A})|\leq |U\setminus A|\leq \aleph_0)$$ olduğundan $$\cup\mathcal{A}\in \tau_2$$ olur.