$\mathbf{T_1})$ $x\in X$ olsun.
$\left.\begin{array}{rr} x\in X \\ \mathbf{(b_1)}\end{array}\right\}\Rightarrow x\in X =\bigcup\mathcal{B}\Rightarrow (\exists B\in\mathcal{B})(x\in B\subseteq X)\Big{/}X\in\tau.$
$[\underset{0}{\underbrace{x\in\emptyset}}\Rightarrow \underset{p}{\underbrace{(\exists B\in\mathcal{B})(x\in B\subseteq \emptyset)}}]\equiv 1.$
O halde $\emptyset\in\tau$ olur.
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$\mathbf{T_2})$ $A_1,A_2\in\tau$ ve $x\in A_1\cap A_2$ olsun.
$\left.\begin{array}{rr} x\in A_1\cap A_2\Rightarrow (x\in A_1)(x\in A_2) \\ A_1,A_2\in\tau\end{array}\right\}\Rightarrow (\exists B_1\in\mathcal{B})(\exists B_2\in\mathcal{B})(x\in B_1\subseteq A_1)(x\in B_2\subseteq A_2)$
$\left.\begin{array}{rr} \Rightarrow (B_1,B_2\in\mathcal{B})(x\in B_1\cap B_2\subseteq A_1\cap A_2) \\ \mathbf{(b_2)}\end{array}\right\}\Rightarrow (\exists B_3\in\mathcal{B})(x\in B_3\subseteq B_1\cap B_2\subseteq A_1\cap A_2).$
O hlade $A_1\cap A_2\in\tau$ olur.
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$\mathbf{T_3})$ $\mathcal{A}\subseteq \tau$ ve $x\in \bigcup \mathcal{A}$ olsun.
$\left.\begin{array}{rr} x\in \bigcup \mathcal{A} \\ \mathcal{A}\subseteq \tau \end{array}\right\}\Rightarrow (\exists A\in\mathcal{A})(x\in A\in \tau)\Rightarrow (\exists B\in\mathcal{B})(x\in B\subseteq A\subseteq\bigcup\mathcal{A})$
O halde $\bigcup\mathcal{A}\in\tau$ olur.