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.

  • $\begingroup$ Have you tried proving that, if $M\in B$, then $M=f(A)$? $\endgroup$ – 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 $f(A)=M$.

  • $\begingroup$ You have talked about a sequence yn and about xi which I am unable to comprehend. Can you please explain that. Thank you. $\endgroup$ – Lawrence Mano Feb 16 '20 at 19:47
  • $\begingroup$ 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. $\endgroup$ – Matematleta Feb 18 '20 at 2:07
  • $\begingroup$ Thank you very much for your exposition $\endgroup$ – Lawrence Mano Feb 18 '20 at 14:04
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$ – Matematleta Feb 18 '20 at 22:23

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