$(X,\tau_1),$ kompakt uzay; $(Y,\tau_2),$ Hausdorff; $f, \ (\tau_1\text{-}\tau_2)$ sürekli ve $f[X]=Y$ olsun.
$$\tau_f:=\{A|(A\subseteq Y)(f^{-1}[A]\in\tau_1)\}=\tau_2$$ olduğunu göstermeliyiz.
$\tau_f\overset{?}{=}\max\{\tau|f, \ (\tau_1\text{-}\tau) \text{ sürekli}\}\Rightarrow \tau_2\subseteq \tau_f\ldots (1)$
$\left.\begin{array}{rr} \underline{A\in\tau_f}\Rightarrow f^{-1}[A]\in\tau_1\Rightarrow f^{-1}[Y\setminus A]=X\setminus f^{-1}[A]\in \mathcal{C}(X,\tau_1) \\ \\ ((X,\tau_1) \text{ kompakt uzay})((Y,\tau_2), \text{ Hausdorff})(f, \ (\tau_1\text{-}\tau_2) \text{ sürekli})\Rightarrow f, (\tau_1\text{-}\tau_2) \text{ kapalı}\end{array}\right\}\Rightarrow$
$\left.\begin{array}{rr} \Rightarrow (f\circ f^{-1})[Y\setminus A]=f[f^{-1}[Y\setminus A]]\in \mathcal{C}(Y,\tau_2) \\ \\ f[X]=Y\Rightarrow f\circ f^{-1}=I_Y\end{array}\right\}\Rightarrow I_Y[Y\setminus A]=Y\setminus A\in \mathcal{C}(Y,\tau_2)$
$\Rightarrow \underline{A\in\tau_2}$
elde edilir yani $$\tau_f\subseteq \tau_2\ldots (2)$$ olur.
$$(1),(2)\Rightarrow \tau_f=\tau_2.$$
O halde $f$ bölüm fonksiyonu.