$\mathbf{1)} \ \emptyset\in\tau^*$ (Tanımdan)
$\left.\begin{array}{rr}(\emptyset \subseteq X)(\emptyset, \ \tau\text{-kompakt})\Rightarrow X\setminus\emptyset\in\tau^*\\ \\ X\setminus \emptyset =X\end{array}\right\}\Rightarrow X\in\tau^*.$
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$\mathbf{2)}$ $A,B\in \tau^*$ olsun.
$\left.\begin{array}{rr}A\in \tau^*\Rightarrow (\exists U\subseteq X)(U, \ \tau\text{-kompakt})(A=X\setminus U) \\ \\ B\in \tau^*\Rightarrow (\exists V\subseteq X)(V, \ \tau\text{-kompakt})(B=X\setminus V)\end{array}\right\}\overset{?}{\Rightarrow}$
$\Rightarrow (U\cup V\subseteq X)(U\cup V, \ \tau\text{-kompakt})(A\cap B=(X\setminus U)\cap (X\setminus V)=X\setminus (U\cup V))$
$\Rightarrow A\cap B\in \tau^*.$
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$\mathbf{3)}$ $\mathcal{A}\subseteq\tau^*$ olsun. $(\emptyset\notin\mathcal{A}$ olduğunu varsaymamızda bir sakınca yok. Neden$?)$
$\left.\begin{array}{rr}\mathcal{B}:=\{B|A\in\mathcal{A}\Rightarrow (\exists B\subseteq X)(B,\ \tau\text{-kompakt})(A=X\setminus B)\} \\ \\ (X,\tau), \text{ Hausdorff }\end{array}\right\}\Rightarrow$
$\overset{(1),(2)}\Rightarrow \left(\bigcap\mathcal{B}, \ \tau\text{-kompakt}\right)\left(\bigcup\mathcal{A}=\bigcup_{A\in\mathcal{A}}A=\bigcup_{B\in\mathcal{B}}\left( X\setminus B\right)=X\setminus \left(\bigcap_{B\in\mathcal{B}} B\right)=X\setminus \left(\bigcap\mathcal{B}\right)\right)$
$\Rightarrow \bigcup\mathcal{A}\in\tau^*.$