Teorem: $A\subseteq\mathbb{R},$ $f\in \mathbb{R}^A$ ve $a\in A$ olmak üzere
$$f, \ a\text{'da sürekli}\Leftrightarrow \left(\forall (x_n)_n\in A^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to f(a))$$
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Kanıt:
$(\Rightarrow):$ $f, \ a\text{'da sürekli},$ $(x_n)_n\in A^{\mathbb{N}}, \ x_n\to a$ ve $\epsilon>0$ olsun.
$\left.\begin{array}{r} \epsilon>0 \\ f, \ a\text{'da sürekli} \end{array} \right\}\Rightarrow \begin{array}{c} \\ \left. \begin{array}{r} (\exists \delta>0)(A\cap (a-\delta,a+\delta)\subseteq f^{-1}[(f(a)-\epsilon,f(a)+\epsilon)]) \\ (x_n\to a)\left((x_n)_n\in A^{\mathbb{N}}\right) \end{array} \right\} \Rightarrow \end{array}$
$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow x_n\in A\cap (a-\delta,a+\delta)\subseteq f^{-1}[(f(a)-\epsilon,f(a)+\epsilon)]) \end{array}$
$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow f(x_n)\in f[A\cap (a-\delta,a+\delta)]\subseteq (f(a)-\epsilon,f(a)+\epsilon)) \end{array}$
$\Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow f(x_n)\in (f(a)-\epsilon,f(a)+\epsilon)).$
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$(\Leftarrow):$ $f, \ a\text{'}$da süreksiz olsun.
$f, \ a\text{'da süreksiz}\Rightarrow (\exists \epsilon>0)(\forall\delta >0)(f[A\cap (a-\delta,a+\delta)]\nsubseteq (f(a)-\epsilon,f(a)+\epsilon))$
$\Rightarrow (\exists \epsilon>0)(\forall n\in\mathbb{N})\left(f\left[A\cap \left(a-\frac1n,a+\frac1n\right)\right]\nsubseteq (f(a)-\epsilon,f(a)+\epsilon)\right)$
$\Rightarrow (\exists \epsilon>0)(\forall n\in\mathbb{N})\left(\exists x_n\in A\cap \left(a-\frac1n,a+\frac1n \right)\right)(f(x_n)\notin (f(a)-\epsilon,f(a)+\epsilon)$
$\Rightarrow \left(\exists (x_n)_n\in A^{\mathbb{N}}\right)(x_n\to a\wedge f(x_n)\nrightarrow f(a)).$
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NOT:
$$\left[\left(\forall (x_n)_n\in A^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to f(a))\right] \Rightarrow \left[f, \ a\text{'da sürekli}\right]$$
$$\equiv$$
$$\left[f, \ a\text{'da sürekli}\right]'\Rightarrow \left[\left(\forall (x_n)_n\in A^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to f(a))\right]'$$
$$\equiv$$
$$f, \ a\text{'da süreksiz}\Rightarrow \left(\exists (x_n)_n\in A^{\mathbb{N}}\right)(x_n\to a \wedge f(x_n)\nrightarrow f(a))$$
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