# Equicontinuity of a Subset of C(X)

Recall from The Set of Real-Valued Continuous Functions on a Compact Metric Space X, C(X) page that if $(X, d)$ is a compact metric space the set $C(X)$ is defined to be the set of all real-valued continuous functions on $X$, that is:

(1)We then defined a very important metric $\rho : C(X) \times C(X) \to [0, \infty)$ given for all $f, g \in C(X)$ by:

(2)We verified that $\rho$ was indeed a metric and so $(C(X), \rho)$ is a metric space.

We will now define another very important concept regarding the set $C(X)$ which is referred to as equicontinuity.

Definition: Let $\Gamma \subseteq C(X)$. Then $\Gamma$ is said to be Equicontinuous on $X$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $f \in \Gamma$ and for all $x, y \in X$ with $d(x, y) < \delta$ we have that $\mid f(x) - f(y) \mid < \epsilon$. |

For example, consider the compact metric space $([0, 1], d)$ where $d$ is the usual Euclidean metric defined for all $x, y \in [0, 1]$ by:

(3)Consider the following subset $\Gamma$ of continuous functions on $[0, 1]$:

(4)We claim that $\Gamma$ is equicontinuous on $[0, 1]$. To show this, let $f_r(x) = r \in \Gamma$ where $r \in \mathbb{R}$. Then we have that for all $\epsilon > 0$ that for any $\delta > 0$ that if $d(x, y) = \mid x - y \mid < \delta$ and for all $x, y \in [0, 1]$ we have that:

(5)We can choose any $\delta$ to satisfy the inequality above for each function $f_r$, and so $\Gamma$ is equicontinuous.