# Monotone Real Function with Everywhere Dense Image is Continuous

## Theorem

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

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

Let $f \left[{ I }\right]$ be everywhere dense in $J$, where $f \left[{ I }\right]$ denotes the image of $I$ under $f$.

Then $f$ is continuous on $I$.

## Proof

Suppose $f$ is increasing.

Suppose $f$ is discontinuous at $c \in I$.

Let $L^{-}$, $L$, and $L^{+}$ denote $\displaystyle \lim_{ x \to c^{-} } f \left({ x }\right)$, $f \left({ c }\right)$, and $\displaystyle \lim_{ x \to c^{+} } f \left({ x }\right)$, respectively.

From Discontinuity of Monotonic Function is Jump Discontinuity, $L^{-}$ and $L^{+}$ are finite.

Since $f$ is increasing:

$L^{-} \leq L \leq L^{+}$

Let $S = \left({ L^{-}, \,.\,.\, L^{+} }\right)$

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, $S \setminus \left\{ { L } \right\}$ is non-empty.

From Finite Subset of Metric Space is Closed, $\left\{ { L } \right\}$ is closed.

From Open Set minus Closed Set is Open, $S \setminus \left\{ { L } \right\}$ is open.

### Lemma

$\displaystyle \left({ \lim_{ x \to c^{-} } f \left({ x }\right) \,.\,.\, \lim_{ x \to c^{+} } f \left({ x }\right) }\right) \cap f \left[{ I }\right] \subseteq \left\{ { f \left({ c }\right) } \right\}$

$\Box$

From the Lemma:

$S \cup f \left[{ I }\right] \subseteq \left\{ { L } \right\}$

Thus:

 $\displaystyle S \cap f \left[{ I }\right] \subseteq \left\{ { L } \right\} \ \$ $\displaystyle \implies \ \$ $\displaystyle \left({ S \cap f \left[{ I }\right] }\right) \setminus \left\{ { L } \right\}$ $\subseteq$ $\displaystyle \left\{ { L } \right\} \setminus \left\{ { L } \right\}$ Set Difference over Subset $\displaystyle$ $=$ $\displaystyle \varnothing$ Set Difference with Self is Empty Set $\displaystyle \implies \ \$ $\displaystyle \left({ S \cap f \left[{ I }\right] }\right) \setminus \left\{ { L } \right\}$ $=$ $\displaystyle \varnothing$ Subset of Empty Set iff Empty $\displaystyle \implies \ \$ $\displaystyle \varnothing$ $=$ $\displaystyle \left({ S \cap f \left[{ I }\right] \setminus \left\{ { L } \right\} }\right)$ Equality is Symmetric $\displaystyle$ $=$ $\displaystyle \left({ S \setminus \left\{ { L } \right\} } \right) \cap f \left[{ I }\right]$ Intersection with Set Difference is Set Difference with Intersection $\displaystyle$ $=$ $\displaystyle \left({ \left({ \left({ L^{-}, \,.\,.\, L }\right) \cup \left\{ { L } \right\} \cup \left({ L, \,.\,.\, L^{+} }\right) }\right) \setminus L }\right) \cap f\left[{ I }\right]$ Break $S$ into $3$ disjoint open intervals $\displaystyle$ $=$ $\displaystyle \left({ \left({ \left({ L^{-}, \,.\,.\, L }\right) \cup \left({ L, \,.\,.\, L^{+} }\right) }\right) \setminus L }\right) \cap f\left[{ I }\right]$ Set Difference with Union is Set Difference $\displaystyle$ $=$ $\displaystyle \left({ \left({ L^{-}, \,.\,.\, L }\right) \cup \left({ L, \,.\,.\, L^{+} }\right) }\right) \cap f\left[{ I }\right]$ Set Difference with Disjoint Set

Thus, there exists a non-empty open subset of $f \left[{ I }\right]$ that is disjoint from $Y$.

From Open Set Characterization of Denseness, $f \left[{ I }\right]$ is not everywhere dense in $J$.

This contradicts the hypothesis that $f \left[{ I }\right]$ 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.

$\blacksquare$