Surjective Monotone Function is Continuous

Theorem
Let $X$ be an open subset of $\R$.

Let $Y$ be a real interval.

Let $f: X \to Y$ be a surjective monotone real function.

Then $f$ is continuous on $X$.

Proof
, let $f$ be strictly increasing.

$f$ is not continuous on $X$.

Then there exists at least one discontinuity at some element of $X$.

It is known that the only discontinuities of a strictly monotonic map are either removable or of the jump type.

It is also known that the one sided limits at a discontinuity can be expressed as either least upper or greatest lower bounds depending on the side being approached.

For each $c \in X$ define:

Suppose that there is jump discontinuity at $c \in X$.

Then:
 * $L^-_c < L^+_c$

By surjectivity there would exist some $a$ such that $L^-_c < \map f a < L^+_c$.

If $a < c$ then $\map f a \le L^-_c$ which contradicts the previous inequality.

There is a similar contradiction if $a \ge c$.

Suppose that there is removable discontinuity at $c \in X$.

Then $\ds L = \lim_{x \mathop \to c} \map f x$ exists, but $L \ne \map f c$.

Hence:
 * $\exists \epsilon \in \R_{>0}: \epsilon = \size {L - \map f c}$

suppose $L > \map f c$.

Then there is a deleted neighborhood $U$ of $c$ such that in $U$:


 * $\size {\map f x - L} < \epsilon$

Hence:
 * $L - \map f x < L - \map f c$

and so:
 * $\map f c < \map f x$

Because $x < c$ this contradicts the fact that $f$ is increasing.

Finally, if $\map f c < L$, then there is a similar contradiction.

It is seen that each case leads to a contradiction

Hence, by Proof by Contradiction, it follows that $f$ can have no discontinuity on $X$.

The result follows.