Gerek Kısmı: d_1\overset{T}{\sim} d_2, \epsilon>0 ve x\in X olsun.
\left.\begin{array}{rr}d_1\overset{T}{\sim} d_2\Rightarrow \tau_{d_1}=\tau_{d_2} \\ \\ (\epsilon>0)(x\in X)\Rightarrow B^{d_1}(x,\epsilon)\in \tau_{d_1} \end{array}\right\}\Rightarrow x\in B^{d_1}(x,\epsilon)\in \tau_{d_2}
\Rightarrow (\exists \delta>0)(B^{d_2}(x,\delta)\subseteq B^{d_1}(x,\epsilon)).
Benzer şekilde
\left.\begin{array}{rr}d_1\overset{T}{\sim} d_2\Rightarrow \tau_{d_1}=\tau_{d_2} \\ \\ (\epsilon>0)(x\in X)\Rightarrow B^{d_2}(x,\epsilon)\in \tau_{d_2} \end{array}\right\}\Rightarrow x\in B^{d_2}(x,\epsilon)\in \tau_{d_1}
\Rightarrow (\exists \delta>0)(B^{d_1}(x,\delta)\subseteq B^{d_2}(x,\epsilon)).
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Yeter Kısmı: d_1\overset{T}{\sim} d_2 olduğunu göstermek için \tau_{d_1}=\tau_{d_2} olduğunu göstermeliyiz. Bunun için de \tau_{d_1}\subseteq \tau_{d_2} ve \tau_{d_2}\subseteq \tau_{d_1} olduğunu göstermek gerekli ve yeterli olacaktır.
\left.\begin{array}{rr} A\in \tau_{d_1}\Rightarrow (\forall a\in A)(\exists \epsilon>0)(B^{d_1}(a,\epsilon)\subseteq A) \\ \\ \text{Hipotez} \end{array}\right\}\Rightarrow
\Rightarrow (\forall a\in A)(\exists \delta>0)(B^{d_2}(a,\delta)\subseteq B^{d_1}(a,\epsilon)\subseteq A)
\Rightarrow A\in \tau_{d_2}
O halde \tau_{d_1}\subseteq \tau_{d_2}\ldots (1)
d_1\overset{T}{\sim} d_2 ve A\in \tau_{d_2} olsun.
\left.\begin{array}{rr} A\in \tau_{d_2}\Rightarrow (\forall a\in A)(\exists \epsilon>0)(B^{d_2}(a,\epsilon)\subseteq A) \\ \\ \text{Hipotez} \end{array}\right\}\Rightarrow
\Rightarrow \forall a\in A)(\exists \delta>0)(B^{d_1}(a,\delta)\subseteq B^{d_2}(a,\epsilon)\subseteq A)
\Rightarrow A\in \tau_{d_1}
O halde \tau_{d_2}\subseteq \tau_{d_1}\ldots (2)
(1),(2)\Rightarrow \tau_{d_1}=\tau_{d_2}\Rightarrow d_1\overset{T}{\sim} d_2.