Teorem: $(X,\tau),(Y,\tau')$ topolojik uzaylar$,$ $f\in Y^X,$ $a\in X$ ve $\mathcal{B}(f(a)), \ f(a)$'da yerel baz olsun.
$$f, \ a\text{'da sürekli}\Leftrightarrow (\forall V\in\mathcal{B}(f(a)))(\exists U\in\mathcal{U}(a))(f[U]\subseteq V).$$
Noktasal sürekliliğin yukarıda verilen karakterizasyonundan faydalanarak şöyle bir kanıt verebiliriz:
$a\in X$ olsun. $y\in\mathbb{R}$ olmak üzere $$\mathcal{B}(y)=\{(y-\epsilon,y+\epsilon)|\epsilon>0\}\subseteq\mathcal{N}(y)$$ ailesi, $y$'de bir yerel bazdır.
$((f+g)(a)-\epsilon,(f+g)(a)+\epsilon)\in\mathcal{B}((f+g)(a))$ verilmiş olsun.
$\left.\begin{array}{rr}V_1\in\mathcal{B}(f(a))\Rightarrow (\exists \epsilon>0)\left(V_1=\left(f(a)-\frac{\epsilon}{2},f(a)+\frac{\epsilon}{2}\right)\right) \\ \\ f, \ a\text{'da sürekli}\end{array}\right\}\Rightarrow $
$\Rightarrow (\exists U_1\in\mathcal{U}(a))\left(f[U_1]\subseteq \left(f(a)-\frac{\epsilon}{2},f(a)+\frac{\epsilon}{2}\right)\right) $
$\Rightarrow (\exists U_1\in\mathcal{U}(a))\left(U_1\subseteq f^{-1}\left[\left(f(a)-\frac{\epsilon}{2},f(a)+\frac{\epsilon}{2}\right)\right)\right] $
$\Rightarrow (\exists U_1\in\mathcal{U}(a))(\forall x\in U_1)\left(x\in U_1\Rightarrow x\in f^{-1}\left[\left(f(a)-\frac{\epsilon}{2},f(a)+\frac{\epsilon}{2}\right)\right)\right]$
$\Rightarrow (\exists U_1\in\mathcal{U}(a))(\forall x\in U_1)\left(f(x)\in f[U_1]\Rightarrow f(x)\in \left(f(a)-\frac{\epsilon}{2},f(a)+\frac{\epsilon}{2}\right)\right)$
$\Rightarrow (\exists U_1\in\mathcal{U}(a))(\forall x\in U_1)\left(f(x)\in f[U_1]\Rightarrow |f(x)-f(a)|<\frac{\epsilon}{2} \right)\ldots (1)$
$\left.\begin{array}{rr} V_2\in\mathcal{B}(g(a))\Rightarrow (\exists \epsilon>0)\left(V_2=\left(g(a)-\frac{\epsilon}{2},g(a)+\frac{\epsilon}{2}\right)\right) \\ \\ g, \ a\text{'da sürekli}\end{array}\right\}\Rightarrow$
$\Rightarrow (\exists U_2\in\mathcal{U}(a))\left(g[U_2]\subseteq \left(g(a)-\frac{\epsilon}{2},g(a)+\frac{\epsilon}{2}\right)\right) $
$\Rightarrow (\exists U_2\in\mathcal{U}(a))\left(U_2\subseteq g^{-1}\left[\left(g(a)-\frac{\epsilon}{2},g(a)+\frac{\epsilon}{2}\right)\right)\right] $
$\Rightarrow (\exists U_2\in\mathcal{U}(a))(\forall x\in U_2)\left(x\in U_2\Rightarrow x\in g^{-1}\left[\left(g(a)-\frac{\epsilon}{2},g(a)+\frac{\epsilon}{2}\right)\right)\right]$
$\Rightarrow (\exists U_2\in\mathcal{U}(a))(\forall x\in U_2)\left(g(x)\in g[U_2]\Rightarrow g(x)\in \left(g(a)-\frac{\epsilon}{2},g(a)+\frac{\epsilon}{2}\right)\right)$
$\Rightarrow (\exists U_2\in\mathcal{U}(a))(\forall x\in U_2)\left(g(x)\in g[U_2]\Rightarrow |g(x)-g(a)|<\frac{\epsilon}{2} \right)\ldots (2)$
$(1),(2)\Rightarrow (U:=U_1\cap U_2\in\mathcal{U}(a))(\forall x\in U)((f+g)(x)\in (f+g)[U]\Rightarrow |(f+g)(x)-(f+g)(a)|=|f(x)+g(x)-f(a)-g(a)|\leq |f(x)-f(a)|+|g(x)-g(a)|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon)$
$\Rightarrow (U:=U_1\cap U_2\in\mathcal{U}(a))(\forall x\in U)((f+g)(x)\in (f+g)[U]\Rightarrow (f+g)(x)\in ((f+g)(a)-\epsilon,(f+g)(a)+\epsilon))$
$\Rightarrow (U:=U_1\cap U_2\in\mathcal{U}(a))((f+g)[U]\subseteq ((f+g)(a)-\epsilon,(f+g)(a)+\epsilon).$
O halde $f+g:X\to\mathbb{R}$ fonksiyonu $a$ noktasında süreklidir. $a\in X$ keyfi olduğundan $f+g$ fonksiyonu $(X$'de$)$ sürekli yani $(\tau\text{-}\mathcal{U})$ süreklidir.
Not: $h:=f+g$ yazmak suretiyle daha sade bir şekilde yazılabilir.