# Statement(s) > [!NOTE] Statement 1 (Equivalence of Cauchy uniformity and uniform convergence) > Let $(f_{n}):\Omega \subset \mathbb{R}\to \mathbb{R}$ be a [[Sequences|sequence]] of [[Real Function|real functions]]. Then $f_{n}$ is [[Uniform Convergence of Sequence of Real Functions|uniformly convergent]] iff it is [[Uniformly Cauchy Sequence of Real Functions|uniformly Cauchy]]. # Proof(s) **Proof of statement 1:** $(\implies)$ Assume $(f_{n})$ is uniformly convergent to $f,$ i.e. for every $\varepsilon>0,$ there exists a natural number $N$ such that for all $n>N,$ $\lvert \lvert f_{n}-f \rvert \rvert_{\infty}< \frac{\varepsilon}{2}.$ Then, for all $m,n>N,$ $\lvert \lvert f_{n} - f_{m} \rvert \rvert_{\infty} \leq \lvert \lvert f_{n} - f+f-f_{m}\rvert \rvert_{\infty} \leq \lvert \lvert f_{n} -f \rvert \rvert_{\infty} + \lvert \lvert f_{m} -f \rvert \rvert_{\infty} \leq \frac{\varepsilon}{2}+ \frac{\varepsilon}{2} =\varepsilon. $ $(\implies )$ Conversely, assume $\left(f_n\right)$ is uniformly Cauchy. That means that for every $x,$ $(f_n(x))_{n=1}^\infty$ is a [[Real Cauchy Sequence|real Cauchy sequence]] and so it follows from the [[General Principle of Convergence|general convergence principle]] that there exists $f$ such that $f_{n}$ converges at least pointwise to $f$. Now, we know that given $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that $\left|f_n(x)-f_m(x)\right|<\varepsilon$ for every $x$ and all $n, m>N$. That is$f_m(x)-\varepsilon<f_n(x)<f_m(x)+\varepsilon \text { for all } x, \text { and all } n, m>N(\varepsilon)$ As the left-hand side holds for all $m>N,$ we can take limits as $m$ goes to infinity. We find $f(x)-\varepsilon \leqslant f_n(x) \leqslant f(x)+\varepsilon \text { for all } x, \quad \text { and all } n>N(\varepsilon) \text {. }$from which it follows that $\left|f(x)-f_n(x)\right|< \varepsilon \text { for all } x, \quad \text { and all } n>N(\varepsilon),$which proves the result. $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography