Pointwise Limit of Increasing Functions is Increasing

Theorem
Let $X \subseteq \R$.

Let $\sequence {f_n}$ be a sequence of real functions $X \to \R$ converging pointwise to a function $f : X \to \R$.

Let $f_n$ be a increasing function for each $n$.

Then $f$ is increasing.

Proof
Suppose that $f$ is not increasing.

That is:


 * there exists $x, y \in X$ such that $x < y$ and $\map f x > \map f y$.

Let:


 * $r = \map f x - \map f y > 0$

Since $\sequence {f_n}$ converges to $f$ converging pointwise, the sequence $\sequence {\map {f_n} x}$ converges to $\map f x$.

That is, there exists $N_1 \in \N$ such that:


 * $\size {\map {f_n} x - \map f x} < \dfrac r 3$

for $n > N_1$.

Similarly, $\sequence {\map {f_n} y}$ converges to $\map f y$.

That is, there exists $N_2 \in \N$ such that:


 * $\size {\map {f_n} y - \map f y} < \dfrac r 3$

for $n > N_2$.

Take $n > \max \set {N_1, N_2}$.

Then, we have:


 * $-\dfrac r 3 < \map {f_n} x - \map f x < \dfrac r 3$

and:


 * $-\dfrac r 3 < \map {f_n} y - \map f y < \dfrac r 3$

so that:


 * $-\dfrac {2 r} 3 < \paren {\map {f_n} x - \map f x} - \paren {\map {f_n} y - \map f y} < \dfrac {2 r} 3$

Note that:

Then, we have:


 * $\map {f_n} x - \map {f_n} y > \dfrac r 3 > 0$

But $x < y$, so this contradicts that $f_n$ is increasing.

So $f$ is increasing.