I am working on a proof that $C(\langle a,b \rangle)$ with sup metric is complete metric space and I am stuck at some point while proving.
Because $\mathbb{R}$ is complete, for every $x\in\langle a,b \rangle$ is a sequence $(f_n(x))$ real convergent sequence with limit denoted by $f(x)$. So we got that $(f_n)$ converges pointwise to $f$, that means for every $x\in\langle a,b \rangle$ exists $m_{0,x}$ such that for every $m>m_{0,x}$ $|f_n(x)-f(x)|<\varepsilon$.
To prove that $(f_n)$ converges $f$ uniformly, we use triangle inequality $|f_n(x)-f(x)|\leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$.
The first expression $|f_n(x)-f_m(x)|$ is by Cauchy's property really smaller than $\frac{\varepsilon}{2}$ but how to choose $m_0$ in a way that it doesn't depend on $x\in\langle a,b \rangle$ so $|f_m(x)-f(x)|<\frac{\varepsilon}{2}$ would be true? $m_0$ as maximum of $m_{0,x}$ doesn't work, because it might be $\infty$. How to choose it? Thank you
