A continuous function on a compact set which takes any value finitely many times.

Asked Apr 10 '17 at 15:22

Active Apr 14 '17 at 12:09

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Let $f : [0,1] \to \mathbb{R}$ be continuous function, such that it takes any value finitely many times and $f(0) \neq f(1)$, then does there exist a point $r_{0}$ in the range such that the cardinality of $f^{-1}(r_{0})$ is odd? (Note that the domain is compact.)

I feel that the answer is affirmative but not been able to come up with any concrete argument. I feel this because, as the function is continuous on $[0,1]$, it would attain its maximum and minimum. With out loss of generality, we can assume that the minimum is attained at $0$ and maximum is attained at $1$. In between, $0$ and $1$ it might oscillate any number of times. But the number of times minimum is attained and maximum is attained must have different parity.

edited Apr 11 '17 at 13:36

asked Apr 10 '17 at 15:22

Raghav

Are you assuming that the range of $f$ is all of $\mathbb R$? – Umberto P. Apr 10 '17 at 15:31
Nope. I am not assuming that the function is surjective. – Raghav Apr 10 '17 at 15:32
What does "it takes any value finitely many times" mean? – Umberto P. Apr 10 '17 at 15:33
@UmbertoP. $\forall x, |f^{-1}({f(x)})|<\infty$ – xavierm02 Apr 10 '17 at 15:34
@Umberto P. Nope. I am not assuming that the function is surjective. – Raghav Apr 10 '17 at 15:34
@xavierm02 You are right, about what i mean by "it takes any value finitely many times". And this question is not same as the question you mentioned because, here the domain is compact. – Raghav Apr 10 '17 at 15:36
@Umberto P. Yes, I mean the same, as mentioned by xavierm02. – Raghav Apr 10 '17 at 15:36
@Jose27 It clearly satisfy all the conditions and every value is taken exactly once, that odd number of times. That is not the question. Question is, if at all it satisfy the conditions mentioned, does it assure that there exist a value of a function which is taken odd number of times? – Raghav Apr 10 '17 at 16:03
You mean the answer is affirmative, not assertive. – Ben Grossmann Apr 10 '17 at 16:25
Possible duplicate of this question. An answer of the question solved what you're considering. – Ctoast Apr 10 '17 at 16:30
@UmbertoP. The image of a continuous $f:[0,1]\to \mathbb R$ must be a closed bounded interval. – DanielWainfleet Apr 10 '17 at 17:07
Thank you, I know that. My concern is about the strange wording of the question. – Umberto P. Apr 10 '17 at 17:15
@Toast Would you kindly elaborate? – Raghav Apr 11 '17 at 14:39
@Omnomnomnom Thank you. – Raghav Apr 11 '17 at 16:55
@Raghav GQBT's first answer seems to be a proof for this question. – Ctoast Apr 12 '17 at 18:22
@Toast Ok. Let me think about it. – Raghav Apr 13 '17 at 10:40
@Toast You are right, it gives the answer. – Raghav Apr 15 '17 at 11:00

A continuous function on a compact set which takes any value finitely many times.

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