# Uniformly Continuous Function is Continuous/Real Function/Proof 2

Jump to navigation
Jump to search

## Theorem

Let $I$ be an interval of $\R$.

Let $f: I \to \R$ be a uniformly continuous real function on $I$.

Then $f$ is continuous on $I$.

## Proof

Let $x \in I$.

Let $\epsilon \in \R_{>0}$.

As $f$ is uniformly continuous:

- $\exists \delta \in \R_{>0}: \paren {x, y \in I, \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon}$

Then, for all $y \in I$ such that $\size {x - y} < \delta$:

- $\size {\map f x - \map f y} < \epsilon$

Thus by definition $f$ is continuous at $x$.

As $x$ was arbitrary, $f$ is continuous on all of $I$.

$\blacksquare$