Show that f is uniformly continuous and that $f_n$ is equicontinuous 
$f_n: A \rightarrow \mathbb{R}$,$n \in \mathbb{N}$ is a sequence of functions defined on $A \subseteq \mathbb{R} $. Suppose that $(f_n)$ converges uniformly to $f:  A \rightarrow \mathbb{R}$, and that each $f_n$ is uniformly continuous on $A$.
1.) Can you show that $f$ is uniformly continuous on A?
2.) Can you show that $(f_n)$ is equicontinuous?



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*We are given that$(f_n)$ converges uniformly to $f$. This means that for every $\epsilon>0$ there exists an $N\in \mathbb{N}$, such that $|f_n(x)-f(x)| <\epsilon$ whenever $n \ge N$ and $x \in A$. We have to show that $f$ is uniformly continuous on $A$, which means that for every $\epsilon >0$ there exists a $\delta>0$ such that $|x-y|<\delta$ implies |$f(x)-f(y)|<\epsilon$

*We need to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|f_n(x)-f_n(y)|< \epsilon$ for all $n\in \mathbb{N}$ and $\|x-y|< \delta$ in $A$

For 1, let $\epsilon >0$. Then pick $n$ such that $|f_n(x) - f(x)| < \epsilon/3$ on $A$. By uniform continuity of $f_n$, there exists a $\delta$ such that $|x-y| < \delta \Longrightarrow |f_n(x)-f_n(y)| < \epsilon/3$. Now if $|x-y| < \delta$,
$$ |f(x)-f(y)| = |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y) -f(y)| \leq $$
$$|f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y) -f(y)|  < \epsilon$$
