# If $(f_n)_{n \in \mathbb{N}}$ is continuous on $\mathbb{R}$, and convergences uniformly on $(a,b)$ show that is converges uniformly on $[a,b]$

Let $(f_n)_{n \in \mathbb{N}}$ be continuous on $\mathbb{R}$ and maps onto $\mathbb{R}$. Show that if $(f_n)_{n \in \mathbb{N}}$ converges uniformly on $(a,b)$ then it converges uniformly on $[a,b]$.

My take on was to use the Cauchy uniform criterion. As we have $\forall \epsilon >0, \exists N \in \mathbb{N},\forall n,m \geq N, \forall x \in (a,b), |f_n(x)-f_m(x)|\leq \epsilon$.

And this is equivalent to $\forall \epsilon >0, \exists N \in \mathbb{N}, \forall n, m \geq N, ||f_n - f_m||_{\infty}^{(a,b)} \leq \epsilon.$

And to prove the statement, my final argument is that thanks to the continuity of $(f_n)$ on $\mathbb{R}$, we have $||f_n - f_m||_{\infty}^{[a,b]} = ||f_n - f_m||_{\infty}^{(a,b)}$, thus we have uniform convergence on $[a,b]$.

Is my argument correct?

Yes, that is correct. More generally, you could replace $(a,b)$ by some subset of $[a,b]$ which is dense on $[a,b]$ (such as $[a,b]\cap\mathbb Q$, for instance).