Analysis Question on Continuity

[dhong]

Homework Statement

Suppose the function $f:[0,1] \to \mathbb{R}$ is continuous, $f(0) > 0$ and $f(1)=0$. Prove that there is a number $x_0 \in (0,1] : f(x_0) = 0$ and $f(x) > 0$ for $0 \leq x \leq x_0$.

Homework Equations

We can't use the IVT. Additionally, the definition of continuity we have been given is, a function $f: D \to \mathbb{R}$ is continuous at $x_0$ in $D$ if whenever $\{x_n \}$ is a sequence in $D$ that converges to $x_0$ the image sequence $\{ f(x_n) \}$ converges to $f(x_0)$.

The Attempt at a Solution

I was thinking about considering the set $S=\{f([0,1]) \}$ and using the Completeness Axiom to label the $\inf S$ then using the Bolzano-Weierstrass Theorem as part of a constructive existence proof.

Can you tell me if I'm on the right track, or is there a better way to start?

Thanks a ton!

 

[LCKurtz]

Homework Statement

Suppose the function $f:[0,1] \to \mathbb{R}$ is continuous, $f(0) > 0$ and $f(1)=0$. Prove that there is a number $x_0 \in (0,1] : f(x_0) = 0$ and $f(x) > 0$ for $0 \leq x \leq x_0$.

Homework Equations

We can't use the IVT. Additionally, the definition of continuity we have been given is, a function $f: D \to \mathbb{R}$ is continuous at $x_0$ in $D$ if whenever $\{x_n \}$ is a sequence in $D$ that converges to $x_0$ the image sequence $\{ f(x_n) \}$ converges to $f(x_0)$.

The Attempt at a Solution

I was thinking about considering the set $S=\{f([0,1]) \}$ and using the Completeness Axiom to label the $\inf S$ then using the Bolzano-Weierstrass Theorem as part of a constructive existence proof.

Can you tell me if I'm on the right track, or is there a better way to start?

Thanks a ton!

I would start with $S=\{x\in [0,1]: f(x) = 0\}$.

 

[pasmith]

I would start with $S=\{x\in [0,1]: f(x) = 0\}$.

The only drawback there is that resort to the IVT seems necessary to show that f(x) > 0 when x < \inf S.

I would suggest instead P = \{ x \in [0,1] : \mbox{$f$ is strictly positive on $[0,x)$} \}, and all that is necessary is to show that f(\sup P) = 0.

 

[LCKurtz]

The only drawback there is that resort to the IVT seems necessary to show that f(x) > 0 when x < \inf S.

True enough. I didn't notice until after I read your post that he can't use the IVT.

 

[exclamationmarkX10]

The only drawback there is that resort to the IVT seems necessary to show that f(x) > 0 when x < \inf S.

I would suggest instead P = \{ x \in [0,1] : \mbox{$f$ is strictly positive on $[0,x)$} \}, and all that is necessary is to show that f(\sup P) = 0.

Don't you mean f(\sup P) > 0?

Another route you could take to prove the result is by contradiction: Suppose that there does not exist x_0 \in (0,1] such that f(x_0) = 0 and f(x) > 0\ \forall\ x \in\ [0, x_0]. That would mean that for all x_0 \in (0,1], f(x_0) \neq\ 0 or f(x) \leq\ 0 for some x \in\ [0, x_0]. Construct a sequence: x_1, x_2, \ldots such that it converges to 0 but the sequence f(x_1), f(x_2), \dots does not converge to a positive number, contradicting the fact that f is continuous.

 

[pasmith]

Don't you mean f(\sup P) > 0?

No, f(\sup P) = 0. I'd explain why f(\sup P) > 0 is impossible, but that would be doing half of the OP's work for him.

 

[exclamationmarkX10]

No, f(\sup P) = 0. I'd explain why f(\sup P) > 0 is impossible, but that would be doing half of the OP's work for him.

My mistake, I thought you were saying \sup P=0