Does there exist a continuous function f:[0,1]-->[0,1] such that for all y in [0,1], $\displaystyle f^{-1}$(y) contains a finite even number (possibly zero) of points in [0,1]. If no, prove it. If yes, construct an example.

Any ideas on this?

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- Dec 5th 2009, 11:38 AMpaupsersDoes this function exist?
Does there exist a continuous function f:[0,1]-->[0,1] such that for all y in [0,1], $\displaystyle f^{-1}$(y) contains a finite even number (possibly zero) of points in [0,1]. If no, prove it. If yes, construct an example.

Any ideas on this? - Dec 6th 2009, 01:02 PMpaupsers
I'm thinking the answer for this is "no" since for it to be continuous there can be no jumps, and therefore if there are no jumps then it must contain infinitely many points in both y and x. Has anyone ever heard of a problem like this before?

- Dec 6th 2009, 01:05 PMpaupsers
Because if the function output a finite even number of values (say, 2) then $\displaystyle f^{-1}(y)$ would only have 2 values to choose from, thus giving only 2 x's. But, if f output only 2 values, then it would be discontinuous. Does that sound right?

- Dec 6th 2009, 01:54 PMredsoxfan325
Yes, I believe if you apply the IVT to $\displaystyle f^{-1}$, it should follow that it contains uncountably many points.

- Dec 6th 2009, 02:00 PMpaupsers
Here's what I'm thinking... Consider the graph of $\displaystyle x^2$ shifted to the right .5 units. Then $\displaystyle f^{-1}(.5)$ has two values, $\displaystyle f^{-1}(.3)$, $\displaystyle f^{-1}(1)$ (where the values are the y-values, by the way), but $\displaystyle f^{-1}(0)$ has only one value. Thus, any continuous graph on [a,b] has a max/min, which would make the value odd.

- Dec 6th 2009, 02:22 PMredsoxfan325
I misunderstood what you were asking. I thought you were referring to the entire inverse image of $\displaystyle f$, but I see now that you a talking about the fibers of individual points.

You seem to be on the right track with the max/min argument, but I think you'd have to prove that if $\displaystyle |f^{-1}(\max(f))|$ is even, then $\displaystyle |f^{-1}(\min(f))|$ is necessarily odd, and vice-versa. - Dec 7th 2009, 09:41 AMLaurent
- Dec 7th 2009, 11:42 AMpaupsers
Well my professor told my class today that this function DOES exist. I was almost positive it didn't. He did say it's extremely tricky, but I think that was obvious. I've been trying all day to find a function (even draw one) that fits the criteria.

Basically, the idea is that if you draw a horizontal line through the function, it will intersect the graph either a finitely even number of times or zero times, for every y between 0 and 1. And the function MUST be continuous on the closed interval [0,1] for both x and y.

I would really appreciate anyone finding this function! - Dec 7th 2009, 03:56 PMpaupsers
Has anyone heard of something like this, or have any ideas?

My problem is:

If there are an odd number of max's or min's, then that function won't work.

If there is an even number of both max's and min's, then there must be an odd number of segments that connect them (it takes (n-1) segments to join n points), thus you can draw a horizontal line through the segments to disprove it.

My guess is, this is not an elementary function at all. I don't even need the function itself; a picture would work just fine! - Dec 7th 2009, 04:47 PMpaupsers
Does this contradict my professor?

JSTOR: An Error Occurred Setting Your User Cookie - Dec 7th 2009, 07:57 PMJose27
Not really, notice that in that paper they talk about functions that are k-to-1 or at most k-to-1. In your problem you have no such restrictions, which should be a hint as to how the function must behave: you can pick a point whose inverse image is as big as you want ie. there is no bound for the size of the inverse images.

Apart from that I'm afraid I haven't really advanced much in this question. - Dec 7th 2009, 08:24 PMpaupsers
I appreciate your input nonetheless.

I'd really love to see the function that satisfies this! - Dec 9th 2009, 07:21 AMDiegoSpivak
This seems like a really interesting problem, I wish I had more time to address it. All I can tell you is that in the Calculus from Spivak a similar problem is dealt with in exercise 20 chapter 7, the problem is the following: show that there does not exist a continous function f defined in the reals such that it takes exactly twice each of the values. Obviously this can be proved, though it might not be that easy, IVT theorem is required to prove this.

- Dec 9th 2009, 11:09 AMLaurent
This problem is much easier: suppose f is such a function. Choose a real $\displaystyle a$, and let $\displaystyle b$ be the only other real such that $\displaystyle f(a)=f(b)$. Without loss of generality, we may assume $\displaystyle a<b$.

Since $\displaystyle f$ is continuous on the segment $\displaystyle [a,b]$, we can define $\displaystyle m=\min_{[a,b]} f = f(x_m)$ and $\displaystyle M=\max_{[a,b]} f=f(x_M)$ where $\displaystyle x_m,x_M\in [a,b]$. We can't have $\displaystyle m=M$ (it would imply $\displaystyle f(x)=f(a)$ for all $\displaystyle x\in[a,b]$), hence we have $\displaystyle m<f(a)$ or $\displaystyle M>f(a)$ (or both). For instance, let us assume that $\displaystyle M>f(a)$.

Suppose (by contradiction) that this maximum is reached at two points inside $\displaystyle [a,b]$: $\displaystyle M=f(x_M)=f(x'_M)$ where $\displaystyle a<x_M<x'_M<b$. By the IVT, there exists $\displaystyle x\in(x_M,x'_M)$ such that $\displaystyle f(a)<f(x)\leq M$ (because $\displaystyle f((x_M,x'_M))$ is a non-empty interval with upper bound $\displaystyle M$ (by the IVT), hence it must intersect $\displaystyle [f(a),M]$). But the IVT on $\displaystyle [a,x_M]$ and $\displaystyle [x'_M,b]$ shows that there are at least two others values which are mapped to $\displaystyle f(x)$, contradicting the hypothesis.

Therefore, there is only one $\displaystyle x_M\in[a,b]$ such that $\displaystyle f(x_M)=M$.

By assumption there exists $\displaystyle x'$ such that $\displaystyle f(x')=f(x_M)$, and we must have $\displaystyle x'<a$ or $\displaystyle x'>b$ because of the previous paragraph. Let's say $\displaystyle x'<a$. Choose $\displaystyle y\in (f(a),f(x'))=(f(a),f(x_M))$. The IVT shows that there is $\displaystyle x''\in(x',a)$ with $\displaystyle f(x')=y$. But the IVT on $\displaystyle [a,x_M]$ and $\displaystyle [x_M,b]$ also provides two other points (in these intervals) that are mapped to $\displaystyle y$. This is a contradiction with the assumption. qed. - Dec 10th 2009, 03:04 PMpaupsers
This is the solution to my original question, if anyone was interested.

http://www.math.ufl.edu/~pilyugin/Pics/evenpreimage.pdf