Uniformly cauchy sequences A sequence of functions $f_n$ is said to be uniformly cauchy if $$\forall \varepsilon > 0 \ \exists N > 0 :\forall z , \forall r, s > N:  |f_r(z) - f_s(z)| < \varepsilon$$
How can I show that if a sequence is uniformly cauchy then $f_n$ converge uniformly to some funciton $f$? We can assume that the metric space is complete.
Thanks!
 A: Here is how to avoid the "let $m \rightarrow \infty$" (see above). Let  $f$ be the pointwise limit (see above). If $\epsilon > 0,$ then so is $\epsilon/2  > 0$ hence there exists a natural number $N_{\epsilon/2}$ such that 
$$\forall z, \forall n,m > N_{\epsilon/2}: \left|f_{n}(z)-f_{m}(z)\right|<\dfrac{\epsilon}{2}$$ 
Let $z$ be arbitrary. If $n > N_{\epsilon/2}$, then we have $$ \left|f(z)-f_{n}(z)\right|\leqslant \left|f(z)-f_{m}(z)\right|+ \left|f_{m}(z)-f_{n}(z)\right|$$ This is true for every $m$. In particular $\left|f(z)-f_{m}(z)\right|<\dfrac{\epsilon}{2}$ if $m>l(z)$, where $l(z)$ is a natural number dependent on $z$, because $f$ is the pointwise limit. Hence, if  $m>\max\left \{N_{\epsilon/2},l(z)\right \}$, we get $$\left|f(z)-f_{n}(z)\right| < \epsilon $$ We have thus shown that for every $\epsilon>0$ there exists an $N$, independent of $z$, such that $$\left|f(z)-f_{n}(z)\right| < \epsilon ,\forall n>N,$$ which is exactly what we needed.
A: Assume that we are given a sequence of functions $f_n:Z\to Y$, and that the metric space $Y$ is complete. According to assumption, for each fixed $z\in Z$ the sequence $\bigl(f_n(z)\bigr)_{n\geq0}$ is a Cauchy sequence in  $Y$. It follows that for each $z\in Y$ the limit $$\lim_{n\to\infty} f_n(z)=: f(z)\in Y$$ exists. I claim that the $f_n$ converge uniformly in $Z$ to the function $f:\>Z\to Y$.
Proof. Let an $\epsilon>0$ be given. Then there is an $N\in{\mathbb N}$ such that for all $z\in Z$ we have $$\bigl|f_{n+m}(z)-f_n(z)\bigr|<\epsilon\tag{1}$$ whenever $n>N$ and $m\geq0$. Letting $m\to\infty$ in $(1)$ we conclude that for all $z\in Z$ we have
$$\bigl|f(z)-f_n(z)\bigr|\leq\epsilon\ ,$$
as soon as $n>N$. Since $\epsilon>0$ was arbitray the claim follows.
Here we have made use of the following Lemma: If $a_m\to a\in{\mathbb R}$ $\>(m\to \infty)$ and $a_m\geq 0$ for all $m\geq 1$ then $a\geq 0$.
A: First, we can show that $f_n$ converges pointwisely to some $f$. Then $\exists N_1$ $|f(z)-f_s(z)|<\epsilon/2 $ $ \forall s>N_1$. (1)
Second, let $\epsilon>0$, $\exists N$ s.t. $\forall z$, we have $|f_s(z) - f_r(z)| < \epsilon/2$. (2)
Now, we want to show that $\forall r>N$ $\forall z$, $|f(z)-f_r(z)|<\epsilon$. Then we are done.
Now choose arbitrary $r>N$, arbitrary $z$.
Using (1), we can find $N_1>N$ s.t. $\forall s>N_1$ and $|f(z)-f_s(z)|<\epsilon/2$. Then $-\epsilon/2+f_s(z) < f(z)<\epsilon/2+f_s(z)$.
Using (2), since $s>N_1>N$ and $r>N$, we can use (2). Then $|f_s(z)-f_r(z)|<\epsilon/2$. Then $-\epsilon/2+f_r(z) < f_s(z) <\epsilon/2+f_r(z)$
Then $|f(z)-f_r(z)|<\epsilon$.
