Question: Let $f_n$ be a sequence of bounded functions on a set $S\subset\mathbb R$. Suppose $f_n\to f$ uniformly on $S$. Prove $f_n$ is uniformly Cauchy on $S$.
Attempt: I proved this easily by ignoring the first sentence (i.e. $f_n$ be a sequence of bounded functions). Can someone please tell me what is the importance that first sentence and how would it help me in constructing my proof?
My version: Let $\varepsilon>0$ be arbitrary. Choose $N$ s.t. $\frac1N < \varepsilon$. Let $n,m > N$. Then $$ \begin{align*} |f_n(x) - f_m(x)| &< |f_n(x) - f(x) + f(x) - f_m(x)|\\ &< |f_n(x) - f(x)| + |f_m(x) - f(x)|\\ &< \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon.\end{align*}$$ QED
Thanks for all the advice!
Your proof is invalid, and I questioned the step in your comments. (Also you should be careful about other inequalities).
How do you prove a general sequence in a metric space is Cauchy if it is convergent?
Let $\epsilon > 0$.
Then also $\dfrac \epsilon 2 > 0$.
Because $f_n$ converges uniformly to $f$, $\exists N$ s.t.$ \forall n > N$, $\forall x \in S$, $|f_n(x) - f(x)| < \dfrac \epsilon 2$ and $\forall m > N, \forall x \in S, |f_m(x) - f(x)| < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then $\forall x \in S$: $$|f_n (x) -f_m(x)| \leq |f_n(x) - f(x)| + |f_m(x) - f(x)| < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$$
So $f_n$ converges uniformly to $f$.