User:Keith.U/Sandbox/Lemma

Theorem
Let $I$ and $J$ be 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 $f$.

Let $f$ be discontinuous at $c \in I$.

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

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

Then:
 * $\displaystyle \exists t \in I : f \left({ t }\right) \in \left({ \lim_{ x \to c^{-} } f \left({ x }\right) \,.\,.\, \lim_{ x \to c^{+} } f \left({ x }\right) }\right)$

Case 1 : $t < c$
Suppose that $t < c$.

So we may discard this case.

Case 2 : $t = c$
Suppose that $t = c$.

So the proposition holds in this case.

Case 3 : $c < t$
Suppose that $t > c$.

So we may discard this case.

So $f \left({ t }\right) = c$, by Proof by Cases.

Thus:

Thus: $z \in \displaystyle \left({ \lim_{ x \to c^{-} } f \left({ x }\right) \,.\,.\, \lim_{ x \to c^{+} } f \left({ x }\right) }\right) \cap f \left[{ I }\right] \implies z \in \left\{ { f \left({ c }\right) } \right\}$

Hence the result, by definition of subset.