Is there a smallest point in the interval [0,1] where f attains the value of 0?

[cxc001]

Suppose f:[0,1]->R is continuous, f(0)>0, f(1)=0.
Prove that there is a X0 in (0,1] such that f(Xo)=0 & f(X) >0 for 0<=X<Xo (there is a smallest point in the interval [0,1] which f attains 0)

Since f is continuous, then there exist a sequence Xn converges to X0, and f(Xn) converges to f(Xo).
Since 0<=(Xo-1/n)<Xo
Can I just let Xn=Xo-1/n so that 0<=Xn<Xo
So when Xn->Xo, f(Xn)->f(Xo)

I wasn't convinced enough this is the right approach...

 

[ystael]

No, this won't work, because you begin by assuming the existence of the number $$x_0$$, which existence you are required to prove.

I suggest two other approaches, either of which will work.

1. What does the set $$f^{-1}(\{0\})$$ look like, topologically?

2. What would happen if the set of points $$x$$ such that $$f(x) = 0$$ had no smallest element in $$[0,1]$$?