A different notion of continuity using open intervals. If the input to a continuous function varies less than delta, then the output values should vary less than epsilon.

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**Theorem 1** (\(\varepsilon\)-\(\delta\) criterion for continuity). *Let
\(I\subset \mathbb{K}\) and \(f:I\to\mathbb{K}\) be a function. Let \(x_0\in I\). Then the following two
statements are equivalent.*

*\(f\) is continuous in \(x_0\);*

*For all \(\varepsilon>0\)
there exists some \(\delta>0\) such
that for all \(x\in I\) with \(|x-x_0|<\delta\) holds \[|f(x)-f(x_0)|<\varepsilon.\]*

*Proof:*

“(i)\(\Rightarrow\)(ii)”: Assume that
(ii) is not fulfilled, i.e., there exists some \(\varepsilon>0\), such that for all \(\delta>0\), there exists some \(x\in I\setminus\{x_{0}\}\) with \(|x-x_0|<\delta\) and \(|f(x)-f(x_0)|>\varepsilon\). As
a consequence, for all \(n\in\mathbb{N}\), there exists some \(x_n\in I\setminus\{x_{0}\}\) with \[|x_0-x_n|<\frac1n\;\text{ and
}\;|f(x_n)-f(x_0)|>\varepsilon.\] Therefore, the sequence
\((x_n)_{n\in\mathbb{N}}\) converges to
\(x_0\), but \(|f(x_n)-f(x_0)|>\varepsilon\), i.e.,
\(f(x_n)\) is not converging to \(f(x_0)\).

“(ii)\(\Rightarrow\)(i)”: Let \((x_n)_{n\in\mathbb{N}}\) be a sequence in
\(I\backslash\{x_0\}\) that converges
to \(x_0\). Let \(\varepsilon>0\). Then there exists some
\(\delta>0\) such that for all \(x\in I\) with \(|x-x_0|<\delta\) holds \(|f(x)-f(x_0)|<\varepsilon\). Since \((x_n)_{n\in\mathbb{N}}\) converges to \(x_0\), there exists some \(N\) such that for all \(n\geq N\) holds \(|x_n-x_0|<\delta\). By the \(\varepsilon\)-\(\delta\)-criterion, we have then for all
\(n\geq N\) that \[|f(x_n)-f(x_0)|<\varepsilon.\] Hence,
\((f(x_n))_{n\in\mathbb{N}}\) converges
to \(f(x_0)\).\(\Box\)

Fabian Gabel and Marcus Waurick. *Well-defined & Wonderful podcast*, marcus-waurick.de.

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