# Statement(s) > [!NOTE] Statement 1 (Uniform Limit of Continuous Functions is Continuous) > Let $\left(f_n\right):\Omega \subseteq \mathbb{R}\to \mathbb{R}$ be a sequence of [[Continuous Real Function|continuous real functions]] that [[Uniform Convergence of Sequence of Real Functions|converge uniformly]] to $f: \Omega \rightarrow R$. Then $f$ is also continuous. # Proof(s) **Proof of statement 1:** By definition, uniform convergence implies that for any $\varepsilon>0,$ there exists some positive integer $N$ such that for all $n>N,$ $\lvert \lvert f_{n} -f\rvert \rvert_{\infty}< \varepsilon/3.$ NTS that given $x_{0}\in \Omega$ and $\varepsilon>0,$ there exist $\delta>0$ such that for all $x\in\left(x_0-\delta, x_0+\delta\right)\cap \Omega,$ we have $\left|f(x)-f\left(x_0\right)\right|<\varepsilon$. With $N$ as above, choose $n>N$, fixed from now own. Since $f_n$ is continuous at $x_0,$ we know that there exists $\delta>0$ such that for all $x \in\left(x_0-\delta, x_0+\delta\right) \cap \Omega$ we have $\left|f_n(x)-f_n\left(x_0\right)\right|<\varepsilon / 3$. Applying the triangle inequality$\begin{aligned} \left|f(x)-f\left(x_0\right)\right|&=\left|f(x)-f_n(x)+f_n(x)-f_n\left(x_0\right)+f_n\left(x_0\right)-f\left(x_0\right)\right| \\ & \leq\left|f(x)-f_n(x)\right|+\left|f_n(x)-f_n\left(x_0\right)\right|+\left|f_n\left(x_0\right)-f\left(x_0\right)\right| \\ & \leq \lvert \lvert f_{n}-f \rvert \rvert_{\infty} + |f_{n}(x)-f_{n}(x_{0})| +\lvert \lvert f_{n}-f \rvert \rvert_{\infty} \leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} +\frac{\varepsilon}{3} \end{aligned}$for any $n>N$ and $x \in\left(x_0-\delta, x_0+\delta\right) \cap \Omega$, with $N$ and $\delta$ chosen as above. $\blacksquare$