Function hunting

Taking $f(x) = x$ with any $A \subset \mathbb{R}$ will trivially produce a true case. Taking bounded continuous $f(x)$ on $\mathbb{R} - \{ 0 \}$ such that $f' > 0$ and $\displaystyle \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$ , produces a false case. It is easy to construct such a function, e.g. $f(x) = \left \{ \begin{matrix} x, x<0 \\ x+1, x>0 \end{matrix} \right.$ . General idea of proof is as follows, to prove that such an $f$ produces a false case:

Suppose not. First, $f' > 0$ on $\mathscr{D}$ , so $f$ qualifies. Then there exists $c \in \mathbb{R}$ by boundedness of $ f $ and continuous $g: \mathbb{R} \to \mathbb{R}$ such that $g (x) = \left \{ \begin{matrix} c, x = 0 \\ f(x), x \in \mathbb{R} - \{ 0 \} \end{matrix} \right.$ , and a $\varepsilon$ without a $\delta \in \mathbb{R}^+$ with the property that whenever $x \in (-\delta, +\delta), |f(x)- c| < \varepsilon$ , contradicting the continuity of $g$ on $\mathbb{R}$ . $\square$