# A continuous function which is strictly increasing or decreasing in closed intervals with infinite pre images at points in its range.

Let $$\lbrace a_{n}\rbrace_{n=1}^{\infty}$$ be a strictly increasing bounded sequence of real numbers such that $$\lim\limits_{n \to \infty}$$ $$a_{n}=A$$. Let $$f:[a_{1},A]\rightarrow \mathbb{R}$$ be a continuous function such that for each positive integer $$i$$, $$f\vert_{[a_{i},a_{i+1}]}:[a_{i},a_{i+1}] \rightarrow \mathbb{R}$$ is either strictly increasing or strictly decreasing. \       Consider the set\       $$B = \left\lbrace M \in \mathbb{R} \middle|\text{ there exist infinitely many }x \in [a_{1},A]\text{ such that } f(x)=M\right\rbrace.$$ Then prove that the cardinality of $$B$$ is atmost one.

If the function is strictly increasing throughout its domain then it is injective and hence $$B$$ is empty. Similarly if it’s decreasing throughout. To have infinite pre images for a single point, the function must have infinite bumps and geometrically it looks that at most one such point is possible. How do we prove it rigorously? Please help.

• Have you tried proving that, if $M\in B$, then $M=f(A)$? – bof Feb 16 '20 at 0:56

Hint: if $$M\in B$$ then there is a sequence $$(y_n)\subseteq [a_1,A]$$ such that $$y_i\neq y_j$$ for all $$i\neq j$$ and such that $$f(x_i)=M$$ for $$i\in \mathbb N$$. Since $$f$$ is injective on each $$[a_i,a_{i+1})$$, each of these subintervals contains at most one $$y_i$$, so there are subsequences $$(a_{n_k})$$ and $$(y_{n_k})$$ such that $$a_{n_k}\le y_{n_k}\le a_{n_{k+1}}$$, from which it follows that $$y_{n_k}\to A$$. Continuity of $$f$$ implies that $$f(A)=M$$.
• If $M\in B$ then there are infinitely many $x\in [a_1,A]$ s.t.$f(x)=M$ But since $f$ is injective on $I_i=[a_i,a_{i+1})$ no two $x's$ can be in any one $I_i$. Now, since $M\in B$ there are infinitely many $x's$ so there must be at least countably many, say $x_k$. And since no two $x_k$ can be in any one $I_i$, we obtain the subsequences of my answer. – Matematleta Feb 18 '20 at 2:07