Why is that hard? Maybe I'm not understanding the requirements. I presume tangent would count as passing through zero times.
Tell me what's wrong with for k an integer and maybe we can understand the task.
The idea of this "proof" is very easy to understand.
Let f:[0,1]-->[0,1] be a continuous function. So you have a square one unit across and one unit tall. Draw a function inside of this box so that if you draw a horizontal line through the function anywhere, it will pass through the function 0 times or an even number of times (not infinitely many times).
Remember, the function has to be continuous on [0,1].
I thought for a while that this function doesn't exist, but my professor today confirmed it DOES exist. Has anyone ever heard of a theorem like this, and/or does anyone know a function that satisfies the criteria?
Take k=1, for example.
Then
So,
In other words, the horizontal line y=1 only intersects the graph of f(x) one time, which is an odd number of times. So that function doesn't work. I need a function so that any horizontal line will cross it finitely many even times or zero times on [0,1], [0,1].
I don't know what you meant about tangent being "zero times."
Do you get what I mean now?
This is the solution, if anyone was interested.
http://www.math.ufl.edu/~pilyugin/Pics/evenpreimage.pdf