# Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 2

## Theorem

Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \R$ be a continuous function.

Then $f$ is uniformly continuous on $\closedint a b$.

## Proof

Aiming for a contradiction, suppose $f$ is not uniformly continuous.

Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:

- $\size {x - y} < \delta$

and:

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

We will now choose for each $k \in \N$ numbers $x_k, y_k \in \closedint a b$ such that:

- $\size {x_k - y_k} < \dfrac 1 k$

and:

- $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$

We have that the sequence $\sequence {x_k}$ is bounded in $\closedint a b$.

So, by the Bolzano-Weierstrass Theorem there exists a convergent subsequence $\sequence {x_{k_j} }$ whose limit lies in $\closedint a b$.

Let this limit be denoted $x_0$.

Now we have:

- $\size {x_0 - y_{k_j} } \le \size {x_0 - x_{k_j} } + \size {x_{k_j} - y_{k_j} } \le \size {x_0 - x_{k_j} } + \dfrac 1 {k_j}$

Therefore the sequence $\sequence {y_{k_j} }$ also converges to $x_0$.

This article, or a section of it, needs explaining.In particular: Make it clear that the above is a definition of sequential continuity in order for the next statement to followIn fact, we should eb able to make use of the fact that a continuous function is already proven to be sequentially continuous, and if it does not so exist, add the above as a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Because $f$ is sequentially continuous at $x_0 \in \closedint a b$, we have:

- $\ds \lim_{j \mathop \to \infty} \map f {x_{k_j} } = \map f {x_0} = \lim_{j \mathop \to \infty} \map f {y_{k_j} }$

This is however, a contradiction, as $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$ for all $k$ and thus all $k_j$.

Therefore $f$ is uniformly continuous.

$\blacksquare$