Uniformly Cauchy

$u_n(x)$ in $x\in A$ is said to be uniformly Cauchy iff:

$$\forall \epsilon \gt 0 \exists N \in \mathbb{Z}^+ \forall m,n \gt N \forall x \in A; \lvert u_n(x)-u_m(x) \rvert \lt \epsilon$$

If $u_n(x)$ is a sequence of real-valued functions, then: $$

$$u_n(x)\text{ converges uniformly} \iff u_n(x)\text{ is uniformly Cauchy}$$

Proof Hint

To prove $\implies$: $$

• Consider $\lvert u_n(x) - u_m(x) \rvert$
• Introduce $u(x)$ in the inequality
• Split the inequality and and use the definition of uniform convergence

To prove $\impliedby$: $$

• Consider the definition of uniformly Cauchy
• $\mathbb{R}$ is complete which implies the convergence.