Gerek ve yeter koşul dendiğine göre ispatı iki adımda yapacağız.
Gerek Kısmı: $d_1\sim d_2$ ve $A\in \tau_{d_2}$ olsun.
$\left.\begin{array}{rr} A\in \tau_{d_2}\Rightarrow i^{-1}[A]=A \\ \\ d_1\sim d_2\Rightarrow \tau_{d_1}=\tau_{d_2}\end{array} \right\}\Rightarrow i^{-1}[A]\in \tau_{d_1}\Big{/} i, (d_1\mbox{-}d_2) \text{ sürekli}$
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$d_1\sim d_2$ ve $A\in \tau_{d_1}$ olsun.
$\left.\begin{array}{rr} A\in \tau_{d_1}\Rightarrow (i^{-1})^{-1}[A]=i[A]=A \\ \\ d_1\sim d_2\Rightarrow \tau_{d_1}=\tau_{d_2}\end{array} \right\}\Rightarrow (i^{-1})^{-1}[A]\in \tau_{d_2}\Big{/} i^{-1}, (d_2\mbox{-}d_1) \text{ sürekli}$
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Yeter Kısmı: $i, (d_1\mbox{-}d_2)$ sürekli ve $A\in\tau_{d_2} $olsun.
$\left.\begin{array}{r} A\in\tau_{d_2} \\ \\ i, (d_1\mbox{-}d_2) \text{ sürekli} \end{array} \right\}\Rightarrow \begin{array}{c} \\ \\ \left. \begin{array}{r} i^{-1}[A]\in \tau_{d_1} \\ \\ i^{-1}[A]=A\end{array} \right\} \Rightarrow A\in\tau_{d_1}\Big{/} \tau_{d_2}\subseteq \tau_{d_1}\ldots (1)\end{array}$
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$i^{-1}, (d_2\mbox{-}d_1)$ sürekli ve $A\in\tau_{d_1} $olsun.
$\left.\begin{array}{r} A\in\tau_{d_1} \\ \\ i^{-1}, (d_2\mbox{-}d_1) \text{ sürekli} \end{array} \right\}\Rightarrow \begin{array}{c} \\ \\ \left. \begin{array}{c} (i^{-1})^{-1}[A]\in \tau_{d_2} \\ \\ (i^{-1})^{-1}[A]=i[A]=A\end{array} \right\} \Rightarrow A\in\tau_{d_2}\Big{/} \tau_{d_1}\subseteq \tau_{d_2}\ldots (2)\end{array}$
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$$(1),(2)$$$$\Rightarrow$$$$\tau_{d_1}=\tau_{d_2} $$$$\Rightarrow$$$$ d_1\sim d_2.$$