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

Proof

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

${\displaystyle {}\lim _{n\rightarrow \infty }f(x_{n})=f(x)\,}$

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

${\displaystyle {}d(x_{n},x)\leq \delta \,}$

holds. By the choice of ${\displaystyle {}\delta }$ we have

${\displaystyle d(f(x_{n}),f(x))\leq \epsilon {\text{ for all }}n\geq n_{0},}$

so that the image sequence converges to ${\displaystyle {}f(x)}$.

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

${\displaystyle d(x_{n},x)\leq {\frac {1}{n}}{\text{ and with }}d(f(x_{n}),f(x))>\epsilon .}$

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