# Homework Help: Analysis Question on Continuity

1. Nov 7, 2014

### dhong

1. The problem statement, all variables and given/known data
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 $0 \leq x \leq x_0$.

2. Relevant 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 $f(x_0)$.

3. 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!

2. Nov 7, 2014

### LCKurtz

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

Last edited: Nov 7, 2014
3. Nov 8, 2014

### pasmith

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 $f(\sup P) = 0$.

4. Nov 8, 2014

### LCKurtz

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

5. Nov 8, 2014

### exclamationmarkX10

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.

Last edited: Nov 8, 2014
6. Nov 9, 2014

### pasmith

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.

7. Nov 9, 2014

### exclamationmarkX10

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