Helly's selection theorem (For sequence of monotonic functions) Let $\{f_n\}$ be a sequence of monotonically increasing functions on $\mathbb{R}$.
Let $\{f_n\}$ be uniformly bounded on $\mathbb{R}$.
Then, there exists a subsequence $\{f_{n_k}\}$ pointwise convergent to some $f$.
Now, assume $f$ is continuous on $\mathbb{R}$.
Here, I want to prove that $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$.
How do i prove this ?
I have proven that " $\forall \epsilon>0,\exists K\in\mathbb{N}$ such that $k≧K \Rightarrow \forall x\in\mathbb{R}, |f(x)-f_{n_k}(x)||<\epsilon \bigvee f_{n_k}(x) < \inf f + \epsilon \bigvee \sup f - \epsilon < f_{n_k}(x)$ ".
The argument is in the link below.
I don't understand why above statement implies "$f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$". Please explain me how..
Reference ;
http://www.math.umn.edu/~jodeit/course/SP6S06.pdf
Thank you in advance!

It's absolutely my fault that i didn't even read (c) in the link. I extend the theorem in the link and my argument below is going to prove;
"If $K$ is a compact subset of $\mathbb{R}$ and $\{f_n\}$ is a sequence of monotonic functions on $K$ such that $f_n\rightarrow f$ pointwise on $K$, then $f_n\rightarrow f$ uniformly on $K$."
(There may exist both $n,m$ such that $f_n$ is monotonically increasing while $f_m$ is monotonically decreasing)
Pf> Since $K$ is closed in $\mathbb{R}$, complement of $K$ is a disjoint union of at most countable open segments.
Let $K^C=\bigsqcup (a_i,b_i)$.
Define;
$$ g_n(x) = 
\begin{cases}
f_n(x) &\text{if }x\in K \\
\frac{x-a_i}{b_i-a_i}f_n(b_i)+\frac{b_i-x}{b_i-a_i}f_n(a_i) & \text{if }x\in(a_i,b_i)\bigwedge a_i,b_i\in\mathbb{R} \\
f_n(b_i) &\text{if }x\in(a_i,b_i)\bigwedge a_i=-\infty \\
f_n(a_i) &\text{if }x\in(a_i,b_i)\bigwedge b_i=\infty
\end{cases} $$
Then, $g_n$ is monotonic on $\mathbb{R}$ and $g_n\rightarrow g$ pointwise on $\mathbb{R}$ ans $g$ is a continuous extension of $f$.
Let $\alpha=\inf K$ and $\beta=\sup K$.
Then, by the argument in the link, $g_n\rightarrow g$ uniformly on $[\alpha,\beta]$.
Hence, $f_n\rightarrow f$ uniformly on $K$.
