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 $f \left[{ I }\right]$ 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.

$\blacksquare$