Surjective Monotone Function is Continuous

Theorem

Let $X$ be an open set 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

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

Let $c \in X$.

From Limit of Monotone Real Function: Corollary, the one sided limits of monotone functions exist:

\(\ds L^-_c\)

\(=\)

\(\, \ds \lim_{x \mathop \to c^-} \map f x \, \)

\(\, \ds = \, \)

\(\ds \sup_{x \mathop < c} \map f x\)

\(\ds L^+_c\)

\(=\)

\(\, \ds \lim_{x \mathop \to c^+} \map f x \, \)

\(\, \ds = \, \)

\(\ds \inf_{x \mathop > c} \map f x\)

and satisfy:

$L^-_c, L^+_c \in Y$

$L^-_c \le \map f c \le L^+_c$

Suppose that $\ds L = \lim_{x \mathop \to c} \map f x$ exists.

From Limit iff Limits from Left and Right:

$L = L^-_c$

This leads to:

$L \le \map f c$

Similarly:

$L = L^+_c$

which leads to:

$L \ge \map f c$

Hence:

$\ds \lim_{x \mathop \to c} \map f x = \map f c$

proving continuity at $c$.

By assumption, $f$ is increasing.

Suppose $\ds \lim_{x \mathop \to c} \map f x$ does not exist.

Then from Discontinuity of Monotonic Function is Jump Discontinuity, there is a jump discontinuity at $c$.

Aiming for a contradiction, suppose $f$ has a jump discontinuity at $c$.

From Real Numbers are Densely Ordered:

$L^-_c < y < L^+_c$

for some $y \in Y$.

By surjectivity, $y = \map f a$ for some $a \in X$.

Hence:

$L^-_c < \map f a < L^+_c$

If $a < c$ then $\map f a \le L^-_c$.

This contradicts the previous inequality.

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

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