There are numbers c, d, with f(a) < f(x) < f(b) for x in (c,d)

[R_beta.v3]

Homework Statement

If $f$ is continuous on $[a,b]$ and $f(a) < f(b)$. Prove that there are numbers $c, d$ with $a \le c < d \le b$ such that $f(c) = f(a)$ and $f(d) = f(b)$ and if $x \in (c,d)$ then $f(a) < f(x) < f(b)$.

Homework Equations

The Attempt at a Solution

This is what I tried.
By considering the set
$A = \left\{ x : a \le x < b \land f(x) = f(a) \right\}$, which is non-empty and bounded above, so it has a least upper bound $\alpha$, then I showed that $f(\alpha) = f(a)$. And by considering the set $B = \left\{ x : \alpha < x \le b \land f(x) = f(b) \right\}$, which is nonempty and bounded below, so it has a greatest lower bound $\beta$. I showed $f(\beta) = f(b)$. And finally showed if $x \in (\alpha, \beta)$ then $f(a) < f(x) < f(b)$. So letting $c = \alpha$ and $d = \beta$.

 

[LCKurtz]

Homework Statement

If $f$ is continuous on $[a,b]$ and $f(a) < f(b)$. Prove that there are numbers $c, d$ with $a \le c < d \le b$ such that $f(c) = f(a)$ and $f(d) = f(b)$ and if $x \in (c,d)$ then $f(a) < f(x) < f(b)$.

Homework Equations

The Attempt at a Solution

This is what I tried.
By considering the set
$A = \left\{ x : a \le x < b \land f(x) = f(a) \right\}$, which is non-empty and bounded above, so it has a least upper bound $\alpha$, then I showed that $f(\alpha) = f(a)$. And by considering the set $B = \left\{ x : \alpha < x \le b \land f(x) = f(b) \right\}$, which is nonempty and bounded below, so it has a greatest lower bound $\beta$. I showed $f(\beta) = f(b)$. And finally showed if $x \in (\alpha, \beta)$ then $f(a) < f(x) < f(b)$. So letting $c = \alpha$ and $d = \beta$.

But might it not be that $\beta < \alpha$?

 

[R_beta.v3]

But might it not be that $\beta < \alpha$?

Well $B$ is bounded below by $\alpha$, so $\alpha$ is a lower bound of B, and since $\beta$ is the greatest lower bound of $B$, $\alpha \le \beta$

 

[LCKurtz]

Well $B$ is bounded below by $\alpha$, so $\alpha$ is a lower bound of B, and since $\beta$ is the greatest lower bound of $B$, $\alpha \le \beta$

Yes. Sorry, I looked at your definitions of A and B and didn't notice you had $\alpha$ instead of $a$ for the lower limit in B. So I think your argument looks good.