Prove that if the sequence of increasing functions $(f_n)$ converges to $f$, then $f$ is increasing 
Let $a,b\in \Bbb R$, $a<b$ and for $n=1,2,3,\ldots$ let $f_n\colon [a,b]\to \Bbb R$ be an increasing function (i.e. $f_n(x)\leq f_n(y)$ if $x\leq y$). Prove that if the sequence $f_1,f_2,f_3,\ldots$ converges to $f$ then $f$ is increasing, and that if $f$ is continuous then the convergence is uniform.


Let $(f_n)$ be a sequence of increasing functions from $[a,b]$ to $\mathbb{R}$ and $f$ be its pointwise limit.
Choose $x$ and $y$ such that $x\ge y$. There are $\epsilon>0$ and $N,M$ such that $|f_n(x)-f(x)|<\epsilon$ and $|f_m(y)-f(y)|<\epsilon$ whenever $n>N$ and $m>M$. If we choose $k>\min\{N,M\}$  then we have
$$f_k(x)-f(x)<\epsilon$$
$$-f_k(y)+f(y)<\epsilon$$
thus $0\le f_k(x)-f_k(y)< f(x)-f(y)+2\epsilon$. Since we can make $\epsilon$ arbitrarly small, $f(x)\ge f(y)$.
I find it difficult to prove the second part. Any hint please?
 A: Suppose the contrary:
$$\exists \varepsilon > 0 : \forall n \in \mathbb{N}, \exists x=x(n) \in [a,b] , m \ge n : |f_m(x)-f(x)| \ge \varepsilon$$
Note that $x(n)$ must have converging subsequence $y_n \to y$ and due to $f$ being continuous, $f(y_n)$ also converges to $f(y)$. Finally:
$$f(y) = \lim_{n \to \infty} f_n(y)$$
leads to contradiction (get $n$ such that $|f(y)-f_n(y)| < \varepsilon/2$ and $|f(y)-f_n(y)| > \varepsilon/2$, and then use the fact that $f$ is uniformly continuous on $[a,b]$).
A: Note that the second part is false if $f_{n}$ is not increasing.
(Hence, Abstraction's proof must be wrong because it does not use the fact that $f_n$ is monotonic increasing.) The following is a counter-example:
Let $f_{n}:[0,1]\rightarrow\mathbb{R}$ be defined by $f_{n}(x)=1_{(0,\frac{1}{n})}(x)$.
Let $f:[0,1]\rightarrow\mathbb{R}$ be defined by $f(x)=0$. Clearly,
$f_{n}\rightarrow f$ pointwisely and $f$ is continuous and monotonic increasing. However,
$f_{n}(\frac{1}{2n})=1$ for all $n$, so the convergence is not uniform.
Let me present a correct proof below.
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Prove by contradiction. Suppose the contrary that $(f_{n})$ does
not converge uniformly to $f$. Then, there exists $\varepsilon_{0}>0$
such that for each $N$, there exist $n\geq N$ and $x\in[a,b]$ such
that $|f_{n}(x)-f(x)|\geq\varepsilon_{0}$. Take $N=1$, we obtain
$n_{1}$ and $x_{1}\in[a,b]$ such that $|f_{n_{1}}(x_{1})-f(x_{1})|\geq\varepsilon_{0}$.
Take $N=n_{1}+1$, we obtain $n_{2}\geq n_{1}+1$ and $x_{2}\in[a,b]$
such that $|f_{n_{2}}(x_{2})-f(x_{2})|\geq\varepsilon_{0}$. Continue
the process indefinitely (to be formal, we invoke Recursion Theorem which
relies on the Axiom of Countable Choice, but we take it for granted),
we obtain a sequence of positive integers $(n_{k})$, with $n_{1}<n_{2}<\ldots$
and a sequence $(x_{k})$ in $[a,b]$ such that for each $k$, we
have $|f_{n_{k}}(x_{k})-f(x_{k})|\geq\varepsilon_{0}$. By passing
to a suitable subsequence, without loss of generality, we may assume
that $x_{k}\rightarrow c$ for some $c\in[a,b]$. Since $f$ is continuous
on the compact set $[a,b]$, it is uniformly continuous on $[a,b].$
For $\varepsilon_{0}/4$, choose $\delta>0$ such that $|f(x)-f(y)|<\varepsilon_{0}/4$
whenever $|x-y|<\delta$.
Case 1: $c\in(a,b)$. Choose $c_{1},c_{2}\in[a,b]$ such that $c_{1}<c<c_{2}$
and $c_{2}-c_{1}<\delta.$ Choose $K_{1}$ such that $x_{k}\in(c_{1},c_{2})$
whenever $k\geq K_{1}$. Choose $K_{2}$ such that $|f_{n_{k}}(c_{1})-f(c_{1})|<\varepsilon_{0}/4$
and $|f_{n_{k}}(c_{2})-f(c_{2})|<\varepsilon_{0}/4$ whenever $k\geq K_{2}$.
Now let $k=\max(K_{1},K_{2})$, then we have
\begin{eqnarray*}
f_{n_{k}}(x_{k})-f(x_{k}) & \leq & f_{n_{k}}(c_{2})-f(x_{k})\\
 & = & \left[f_{n_{k}}(c_{2})-f(c_{2})\right]+\left[f(c_{2})-f(x_{k})\right]\\
 & < & \frac{\varepsilon_{0}}{4}+\frac{\varepsilon_{0}}{4}\\
 & = & \frac{\varepsilon_{0}}{2}.
\end{eqnarray*}
On the other hand, 
\begin{eqnarray*}
f_{n_{k}}(x_{k})-f(x_{k}) & \geq & f_{n_{k}}(c_{1})-f(x_{k})\\
 & = & \left[f_{n_{k}}(c_{1})-f(c_{1})\right]+\left[f(c_{1})-f(x_{k})\right]\\
 & > & -\frac{\varepsilon_{0}}{4}-\frac{\varepsilon_{0}}{4}\\
 & = & -\frac{\varepsilon_{0}}{2}.
\end{eqnarray*}
Therefore 
$$
\left|f_{n_{k}}(x_{k})-f(x_{k})\right|<\frac{\varepsilon_{0}}{2},
$$
which is a contradiction.
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The cases $c=a$, $c=b$ can be proved similarly. For example,
if $c=b$, we modify the above proof by choosing $c_{1},c_{2}\in[a,b]$
such that $c_{1}<c=c_{2}$ and $c_2-c_1<\delta$.
