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 $X$.

, let $f$ be increasing.

Note: it is not necessary for $f$ to be strictly increasing.

But it is interesting to observe that with the given hypothesis, the interval $Y$ must be an open interval, not an arbitrary interval. For example, $Y$ cannot have a left endpoint because if it did by the surjectivity of $f$  it would of the form $f(a)$, $a \in X$. Since $X$ is open, there exists $b \in X, b < a $, and because $f$ is increasing $f(b) < f(a)$ and that contradicts that $f(a)$ is a left endpoint.

Proof
$f$ is not continuous on $X$.

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

If $f$ is strictly increasing, then the following equations are known, for example see Lebl's online analysis book listed in the sources, Proposition 3.6.2.

From this it follows that the only discontinuities of $f$ are either removable or of the jump type. 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 $a \ge c$. In the case of a removable discontinuity $L = \lim_{x \, \to \, c}f(x) $ exists but is not equal to $ f(c) $. If $L > f(c)$, then $\epsilon = L-f(c)$ is positive and thus there is a deleted neighborhood $U$ of $c$ such that in that neighborhood $| f(x) -L| < \epsilon $ implies $L-f(x) < L-f(c)$ If we take $x \in U$ with $x < c$, this leads to $f(c)<f(x)$. Since $x<c$ this contradicts that $f$ was increasing. Finally, if $f(c)<L$, then there is a similar contradiction.