I am reviewing the proof of the Cauchy Criterion for sequences of functions and have a question regarding the backwards direction.
Statement: Let $A\subseteq \mathbb{R}$ and $(f_n)$ be a sequence of real-valued functions with domain $A$. Then $(f_n)$ converges uniformly if and only if $\forall \epsilon >0$, $\exists N\in \mathbb{N}$ such that $m,n\geq N$ and $x\in A$ implies $|f_n(x)-f_m(x)|<\epsilon$.
My question: For the backwards direction, we know that for each $x$ the sequence $(f_n(x))$ converges (to say $L_x$), but how do we know there is a single $N$ such that $n\geq N$ implies $|(f_n(x)-L_x|, |f_n(y)-L_y|<\epsilon$ for any choice of $x,y\in A$? The difficulty I am having is that the $N$ given in the proof of the Cauchy criterion for real sequences is dependent on the subsequence chosen (I am assuming a proof that uses the BW theorem). An answer is appreciated, but I would prefer a hint at this point to help.
Best Answer
Take $\varepsilon>0$, and take $\varepsilon'\in(0,\varepsilon)$. Then there is a $N\in\Bbb N$ such that$$m,n\geqslant N\implies|f_m(x)-f_n(x)|<\varepsilon'.$$ But then, if $n\geqslant N$,\begin{align}|L_x-f_n(x)|&=\left|\lim_{m\to\infty}f_m(x)-f_n(x)\right|\\&\leqslant\varepsilon'\\&<\varepsilon.\end{align}