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
$\left.\begin{array}{rr}d_1\overset{T}{\sim} d_2\Rightarrow \tau_{d_1}=\tau_{d_2} \\ \\ (\epsilon>0)(x\in X)\Rightarrow B^{d_2}(x,\epsilon)\in \tau_{d_2} \end{array}\right\}\Rightarrow x\in B^{d_2}(x,\epsilon)\in \tau_{d_1}$
$\Rightarrow (\exists \delta>0)(B^{d_1}(x,\delta)\subseteq B^{d_2}(x,\epsilon)).$
$$-----------------------------------$$
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 $\tau_{d_2}\subseteq \tau_{d_1}\ldots (2)$
$$(1),(2)\Rightarrow \tau_{d_1}=\tau_{d_2}\Rightarrow d_1\overset{T}{\sim} d_2.$$