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Let $f:[a,b]\rightarrow[f(a),f(b)]$ be strictly increasing continuous function (i.e $x>y \implies f(x)>f(y)$). Prove that f is invertible.

Proving that the function is one-to-one was simple enough. I need some guidance on proving it's onto. I'm new to $\epsilon-\delta$ proofs since I just started self-studying some metric spaces.

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Apply the Intermediate Value Theorem. –  Vectk Jul 19 '12 at 17:51
    
Fix $d$ in $(f(a),f(b))$ and suppose there is no $c$ in [a,b] such that $f(c)=d$ and try to contradicts continuity using the other hypothesis –  Jean-Sébastien Jul 19 '12 at 17:51
    
Since $[a,b]$ is connected, and $f$ is continuous, then $f([a,b])$ is connected. Connected subsets of the real line are intervals. Since $f$ is monotonic, we have $f([a,b]) = [f(a),f(b)]$. –  copper.hat Jul 19 '12 at 17:53
    

1 Answer 1

up vote 1 down vote accepted

Let $y \in (f(a), f(b))$ and let $A = \{ x \in [a, b] : f(x) < y\}$. Since A is nonempty, and $A \subset [a, b]$, $s =\sup A$ exists. To show that $f(s) = y$, choose a sequence $x_n \in A$ that converges to $s$. Since we have $s \in [a, b]$, and by continuity,

$$f(s) = \lim_{n \to \infty}f(x_n) \le y$$

By contradiction, suppose $f(s) < y$. Then $y - f(x)$ is continuous on $[a, b]$, and $y - f(s) > 0$. We can choose $\delta > 0$ and $u > s$ such that $y-f(u) > \delta$, but this implies that $u \in A$, a contradiction.

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