# 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

Without loss of generality, let $f$ be increasing.

Aiming for a contradiction, suppose $f$ is not continuous on $X$.

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

From Limit of Monotone Real Function: Corollary, the following limits exist.

\(\ds L^-_c\) | \(=\) | \(\ds \lim_{x \mathop \to c^-} \map f x\) | ||||||||||||

\(\ds L^+_c\) | \(=\) | \(\ds \lim_{x \mathop \to c^+} \map f x\) |

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 $\ds L = \lim_{x \mathop \to c} \map f x$ exists but is not equal to $\map f c$.

If $L > \map f c$, then $\epsilon = L - \map f c$ is positive and thus there is a deleted neighborhood $U$ of $c$ such that in that neighborhood:

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

If we take $x \in U$ with $x < c$, this leads to $\map f c < \map f x$.

Since $x < c$ this contradicts that $f$ was increasing.

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

$\blacksquare$

## Tangential and irrelevant digression

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 $\map f a$, $a \in X$. Since $X$ is open, there exists $b \in X, b < a$, and because $f$ is increasing $\map f b < \map f a$ and that contradicts that $\map f a$ is a left endpoint.

## Sources

- 2021: Jiří Lebl:
*Basic Analysis I*: Proposition $3.6.6$, section $3.6$