Monotone Real Function with Everywhere Dense Image is Continuous

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Theorem

Let $I$ and $J$ be real intervals.

Let $f: I \to J$ be a monotone real function.

Let $f \sqbrk I$ be everywhere dense in $J$, where $f \sqbrk I$ denotes the image of $I$ under $f$.


Then $f$ is continuous on $I$.


Proof

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

Aiming for a contradiction, suppose $f$ is discontinuous at $c \in I$.

Let:

$L^-$ denote $\ds \lim_{x \mathop \to c^-} \map f x$
$L$ denote $\map f c$
$L^+$ denote $\ds \lim_{x \mathop \to c^+} \map f x$

From Discontinuity of Monotonic Function is Jump Discontinuity, $L^-$ and $L^+$ are (finite) real numbers.

From Limit of Monotone Real Function: Corollary:

$L^- \le L \le L^+$

Let $S = \openint {L^-} {L^+}$.

If $L^- = L = L^+$, then $c$ is not a discontinuity, a contradiction.

Thus $S$ is non-degenerate.

From Non-Degenerate Real Interval is Infinite, $S$ is infinite.

From Relative Difference between Infinite Set and Finite Set is Infinite, $S \setminus \set L$ is non-empty.

From Finite Subset of Metric Space is Closed, $\set L$ is closed.

From Open Set minus Closed Set is Open, $S \setminus \set L$ is open.


Lemma

$\ds \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I \subseteq \set {\map f c}$

$\Box$


Thus:

\(\ds S \cap f \sqbrk I\) \(\subseteq\) \(\ds \set L\) Lemma
\(\ds \leadsto \ \ \) \(\ds \paren {S \cap f \sqbrk I} \setminus \set L\) \(\subseteq\) \(\ds \set L \setminus \set L\) Set Difference over Subset
\(\ds \) \(=\) \(\ds \O\) Set Difference with Self is Empty Set
\(\ds \leadsto \ \ \) \(\ds \paren {S \cap f \sqbrk I} \setminus \set L\) \(=\) \(\ds \O\) Subset of Empty Set iff Empty
\(\ds \leadsto \ \ \) \(\ds \O\) \(=\) \(\ds \paren {S \cap f \sqbrk I \setminus \set L}\) Equality is Symmetric
\(\ds \) \(=\) \(\ds \paren { S \setminus \set L} \cap f \sqbrk I\) Intersection with Set Difference is Set Difference with Intersection
\(\ds \) \(=\) \(\ds \paren {\paren {\openint {L^-} L \cup \set L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I\) Break $S$ into $3$ disjoint open intervals
\(\ds \) \(=\) \(\ds \paren {\paren {\openint {L^-} L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I\) Set Difference with Union is Set Difference
\(\ds \) \(=\) \(\ds \paren {\openint {L^-} L \cup \openint L {L^+} } \cap f \sqbrk I\) Set Difference with Disjoint Set

Thus there exists a non-empty open subset of $f \sqbrk I$ that is disjoint from $Y$.


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From Open Set Characterization of Denseness, $f \sqbrk I$ is not everywhere dense in $J$.

This contradicts the hypothesis that $f \sqbrk I$ is everywhere dense in $J$.

Therefore $I$ contains no discontinuities, by Proof by Contradiction.


The same argument, mutatis mutandis, proves the case where $f$ is decreasing

Hence the result, by Proof by Cases.

$\blacksquare$

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