Recent content by anon3335

I have solved the problem with this specific way, but I wanted to solve the problem with another approach.

Ah I see. Yes the argument would not hold. Thanks for your help.

Yes you are right. My argument is flawed in that step, but what if we delete that part from the proof? wouldn't the proof still hold?

Yup.

Let ##x_1## ≠ ##x_2##. Then one gets ##f(x_1) ≠ f(x_2)## according to our assumption and so ##f## is injective.

Since then ##f## would be a continuous injective function on ##I## and so it is strictly monotonic.

Homework Statement Question: Let ##I = [0,1]##. Suppose ##f## is a continuous mapping of ##I## into ##I##. Prove that ##f(x) = x## for at least one ##x∈I##. Homework Equations Define first(##[A,B]##) = ##A## and second(##[A,B]##) = ##B## where ##[A,B]## is an interval in ##R##. The Attempt at...