# Statements ###### For real-valued functions on sets of real numbers Let $\left(f_n\right):\Omega \subseteq \mathbb{R}\to \mathbb{R}$ be a sequence of [[Continuous Real Function|continuous real functions]] that [[Convergence|converge]] to $f: \Omega \rightarrow R$ with respect to the [[Supremum norm|supremum norm]]. Then $f$ is also continuous. ###### For real-valued functions on metric spaces Let $(X,d)$ be a non-empty [[Metrics|metric space]] and let $C(X)=\{ f:X\to \mathbb{R} \mid f \text{ is continuous} \}$, denote the set of continuous real-valued on $X$. Then for any sequence $(f_{n})$ in $(C(X), \lVert \cdot \rVert)_{\infty }$ such that $f_{n}\to f$, we have that $f$ is continuous, where $\lVert \cdot \rVert_{\infty}$ denotes the supremum norm. # Proof ###### Proof for real-valued functions on metric spaces By the definition of convergence, for any $\varepsilon>0,$ there exists some positive integer $N$ such that $\lvert \lvert f_{N} -f\rvert \rvert_{\infty}< \varepsilon/3.$ Fix $x\in X$. Since $f_N$ is continuous at $x$, there exists $\delta>0$ such that for all $y\in X$, $d(x,y)< \delta$ implies that $\left|f_N(x)-f_N\left(y\right)\right|<\varepsilon / 3.$from which it follows, by the triangle inequality, that $\begin{aligned} \left|f(x)-f\left(y\right)\right| & \leq\left|f(x)-f_N(x)\right|+\left|f_N(x)-f_N\left(y\right)\right|+\left|f_N\left(y\right)-f\left(y\right)\right| \\ & \leq \lvert \lvert f_{N}-f \rvert \rvert_{\infty} + |f_{N}(x)-f_{N}(y)| +\lvert \lvert f_{N}-f \rvert \rvert_{\infty} \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon, \end{aligned}$which completes the proof.