# A uniformly Cauchy sequence of functions is uniformly convergent proof

Let $$f_n:A\subseteq\mathbb{R}\to\mathbb{R}$$

$$f_n$$ is uniformly Cauchy $$\implies$$ $$\exists f:A\to\mathbb{R}$$ : $$f_n\xrightarrow{u}f$$ in A

proof.

$$\forall\varepsilon>0$$ $$\exists\nu$$ : $$\forall n,m>\nu$$ $$\sup_{x\in A}|f_n(x)-f_m(x)|<\varepsilon$$

I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof

$$\implies \forall\varepsilon>0$$ $$\exists\nu\in\mathbb{N}$$ : $$\forall n,m>\nu$$ $$\forall x\in A$$ $$|f_n(x)-f_m(x)|<\varepsilon$$

Where is sup now?

So if $$m\to+\infty$$

$$\implies \forall\varepsilon>0$$ $$\exists\nu\in\mathbb{N}$$ : $$\forall n>\nu$$ $$\forall x\in A$$ $$|f_n(x)-f(x)|\leq\varepsilon$$

Ok it's clear why m disappears but I don't understand why $$<\varepsilon$$ becomes $$\leq\varepsilon$$

$$\implies \forall\varepsilon>0$$ $$\exists\nu\in\mathbb{N}$$ : $$\forall n>\nu$$ $$\sup_{x\in A}|f_n(x)-f(x)|\leq\varepsilon$$

This is not clear, where does sup come from?

$$\implies\lim_{n\to\infty}sup_{x\in A}|f_n(x)-f(x)|=0$$

This is directly from the definition of limit, and this means that

$$\implies f_n\xrightarrow{u}f$$ in $$A$$

• Please consider accepting my answer if it has helped :) – user667 Nov 19 '18 at 14:11

Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $$v$$ such that $$\sup_{x\in A}|f_n(x)-f_m(x)|<\epsilon$$ whenever $$n,m>v$$ then we are guaranteed that $$\forall x\in A, |f_n(x)-f_m(x)|<\epsilon$$ whenever $$n,m>v$$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $$|f_n(x)-f(x)|<\epsilon$$ for all $$x\in A$$, we have that $$\epsilon$$ is an upper bound over all $$x$$. In particular, it could be the least upper bound (supremum). Thus we have to write $$\sup_{x \in A}|f_n(x)-f|\leq\epsilon$$ since equality could hold.

I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $$x \in A$$. We have, $$|f_n(x)-f(x)|\leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $$f(x)$$ is the pointwise limit of the sequence $$\{f_n(x)\}$$. We know that there is a pointwise limit because $$\mathbb{R}$$ is complete. By definition of pointwise convergence, there exists some $$N_1$$ such that if $$m>N_1$$ we have $$|f_m(x)-f(x)|<\frac{\epsilon}{2}$$. By assumption, we can choose $$N$$ such that if $$n,m>N$$ we have $$|f_n(x)-f_m(x)|<\frac{\epsilon}{2}$$. Now set $$m>\max\{N_1,N\}$$ (this step is a rigorous way of saying $$m \rightarrow \infty$$). Putting it all together $$|f_n(x)-f(x)|\leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$Because $$x$$ was an arbitrary point of $$A$$, we have found an $$N$$ (independent of $$x\in A$$) such that, $$\forall x\in A, n>N, |f_n(x)-f(x)|<\epsilon$$