continuous function?

Question

prove that a continuous real valued function on a closed inteval in E^2

can not be one-one.

It looks like a problem about extreme value theorem

but cannot figure it out.

closed interval in E^2 means closed rectangle on the plane, I guess.
and f: (a,b) --> r is continuous
(a,b) is in E^2 and r in R

Answer 1

Is E any closed interval or is it E=[0,1]?

I think that you are missing part of the question, there are real valued continuous functions that are one-to-one on a closed interval: namely f(x)=x is one-to-one. Maybe you mean a function from E^2 to E^2? Or maybe you are looking at the Brower fixed point theorem.

Answer 2

Don't forget, S is a compact subset of E^2,
so f continuous => f(S) is a compact subset of R => f(S) is [a,b] like subset of R

So f: S -> f(S) is a surjective function

If f were one to one, f would be a bijective function so once chosen c in (a,b) there is one only y in S such that f(y)=c

Let T = S \ {y}

T is still a connected subset and the restriction of f to T is still
continuous so f(T) would be a connected subset.
But c belongs to (a,b) => f(T) = [a,b] \ {c} isn't connected => f cannot be one to one.

Excuse me but I'm not so skilled in English ;-)