Proof about a real-valued continuous function?

[cookiesyum]

Proof about a real-valued continuous function??

Homework Statement

Five line segments meet at a point. Show that any continuous real-valued function defined on this set must take the same value three times.

The Attempt at a Solution

Take the values f(0,0,0) f(0,0,1) f(1,0,0) f(0,1,0) f(-1,0,0) f(-1,0,0). If three of the values are the same, the proof is done.

Assume all of the values are different.

At least three of f(0,0,1) f(1,0,0) f(0,1,0) f(-1,0,0) f(0,-1,0) are greater than or less than f(0,0,0).

Without loss of generality, assume f(-1,0,0) < f(0,0,0) < f(0,1,0) < f(0,0,1).

Then, by the Intermediate Value Theorem, there is a point p1 between f(-1,0,0) and f(0,0,1). Then p would be on the line segment from (-1,0,0) to (0,0,1) and also on two other line segments...but which??

Hence, the real-valued function takes the same value three times.

 

[HallsofIvy]

Homework Statement

Five line segments meet at a point. Show that any continuous real-valued function defined on this set must take the same value three times.

The Attempt at a Solution

Take the values f(0,0,0) f(0,0,1) f(1,0,0) f(0,1,0) f(-1,0,0) f(-1,0,0). If three of the values are the same, the proof is done.

You were given only five line segments. How can you assume these particular points are on the segments? Also are you assuming that one of the points is the point of intersection and the other five are on fived different segments?

Assume all of the values are different.

How can you do that? If three are not the same, it might still be that two are the same, another two different from the first two but the same, the fifth different from the previous.

At least three of f(0,0,1) f(1,0,0) f(0,1,0) f(-1,0,0) f(0,-1,0) are greater than or less than f(0,0,0).

Without loss of generality, assume f(-1,0,0) < f(0,0,0) < f(0,1,0) < f(0,0,1).

You just said "At least three of f(0,0,1) f(1,0,0) f(0,1,0) f(-1,0,0) f(0,-1,0) are greater than or less than f(0,0,0)" which I interpret to mean "at least three above or at least three below". How does it follow that there are two above and one below? Why not all five above?

Then, by the Intermediate Value Theorem, there is a point p1 between f(-1,0,0) and f(0,0,1). Then p would be on the line segment from (-1,0,0) to (0,0,1) and also on two other line segments...but which??

How do you know that the line segment from (-1,0,0) to (0,0,1) is one of the five given line segments? And unless p is the point of intersection of the five line segments, if p is on one segment, it can't be on any other!

Hence, the real-valued function takes the same value three times.

 

[Dick]

For notation let's just call the intersection point p0, and the other five endpoints p1,p2,p3,p4,p5. It'll save you writing a lot of ones and zeros. And pick the indices so that f(p1)<=f(p2)<=f(p3)<=f(p4)<=f(p5). Now consider all of the cases of where the value of f(p0) lies in the order relationship of the other p's.