different between uniformly cauchy and uniformly converge

Question

I am trying to figure out whats the different between "uniformly converge" and "uniformly cauchy" I been thinking if its cauchy, then think about in graph that two lines are getting close,, how can it be uniform???

Thanks

Answer 1

Here are the definitions :

Consider a set $X$, a metric space $(E,d)$ and a sequence of maps $f_n:X\to E$.

• The sequence $(f_n)$ is said to be uniformly convergent if there exists a map $g:X\to E$ such that :

$$\lim_{n\to\infty}\sup_{x\in X}d\left(f_n(x),g(x)\right)=0$$

• The sequence $(f_n)$ is said to be uniformly Cauchy if :

$$\forall\epsilon>0,\exists N\in\mathbb{N};\,\forall p\ge N,\forall q\ge N,\;\sup_{x\in X}d\left(f_p(x),f_q(x)\right)\le\epsilon$$

It can be proved that :

1. without any assumption on $(E,d)$, uniform convergence implies uniform Cauchy criterion.

2. if $(E,d)$ is complete, those two notions are equivalent.

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Proof of UCV $\implies$ U-Cauchy

Given $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $n\ge N$ implies :

$$\forall x\in X,\,d\left(f_n(x),g(x)\right)\le\frac\epsilon2$$

If $p\ge N$ and $q\ge N$ then for all $x\in X$, by triangle inequality :

$$d\left(f_p(x),f_q(x)\right)\le d\left(f_p(x),g(x)\right)+d\left(g(x),f_q(x)\right)\le\frac\epsilon2+\frac\epsilon2=\epsilon$$Hence :

$$\sup_{x\in X}d\left(f_p(x),f_q(x)\right)\le\epsilon$$as desired.$\tag{QED}$