Condition for Continuity on Interval

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Theorem

Let $f$ be a real function defined on an interval $\mathbb I$.


Then $f$ is continuous on $\mathbb I$ if and only if:

$\forall x \in \mathbb I: \forall \epsilon > 0: \exists \delta > 0: y \in \mathbb I \land \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$


Proof

Let $x \in \mathbb I$ such that $x$ is not an end point.

Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ is the same as $\size {x - y} < \delta$ provided $\delta$ is small enough.

The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x} \map f y = \map f x$, that is, that $f$ is continuous at $x$.


Now suppose $x \in \mathbb I$ and $x$ is a left hand end point of $\mathbb I$.

Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ reduces to $x \le y < x + \delta$ provided $\delta$ is small enough.

The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x^+} \map f y = \map f x$, that is, that $f$ is continuous on the right at $x$.

Similarly, if $x \in \mathbb I$ and $x$ is a right hand end point of $\mathbb I$, then the criterion reduces to the statement that $f$ is continuous on the left at $x$.


Thus the assertions are equivalent to the statement that $f$ is continuous at all points in $\mathbb I$, that is, that $f$ is continuous on $\mathbb I$.

$\blacksquare$


Sources

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