# Real Function/Continuity in a point/Characterization/Fact/Proof

Proof

Suppose that (1) is fulfilled and let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be a sequence in ${}D$ converging to ${}x$ . We have to show that

${}\lim _{n\rightarrow \infty }f(x_{n})=f(x)\,$ holds. To show this, let ${}\epsilon >0$ be given. Due to (1) there exists a ${}\delta >0$ fulfilling the estimation property (from the definition of continuity) and because of the convergence of ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ to ${}x$ there exists a natural number ${}n_{0}$ such that for all ${}n\geq n_{0}$ the estimate

${}d(x_{n},x)\leq \delta \,$ holds. By the choice of ${}\delta$ we have

$d(f(x_{n}),f(x))\leq \epsilon {\text{ for all }}n\geq n_{0},$ so that the image sequence converges to ${}f(x)$ .

Suppose now that (2) is fulfilled. We assume that ${}f$ is not continuous. Then there exists an ${}\epsilon >0$ such that for all ${}\delta >0$ there exist elements ${}z\in D$ such that their distance to ${}x$ is at most ${}\delta$ , but such that the distance of their value ${}f(z)$ to ${}f(x)$ is larger than ${}\epsilon$ . This holds in particular for every ${}\delta =1/n$ , ${}n\in \mathbb {N} _{+}$ . This means that for every natural number ${}n\in \mathbb {N} _{+}$ there exists a ${}x_{n}\in D$ with

$d(x_{n},x)\leq {\frac {1}{n}}{\text{ and with }}d(f(x_{n}),f(x))>\epsilon .$ This sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ converges to ${}x$ , but the image sequence does not converge to ${}f(x)$ , since the distance of its members to ${}f(x)$ is always at least ${}\epsilon$ . This contradicts condition (2).