Let $f:[a,b]\to[f(a),f(b)]$ be monotone increasing and continuous. Prove that $f$ is a homeomorphism. (w/o IVT) I'm reading Intro to Topology by Mendelson. 
The section is titled Connectedness of the Real Line, it's the section right before IVT is introduced.
I've been working on this problem, but I'm stuck on how to show that the function is onto. I know that because its monotone (strictly) increasing, that $f$ must be one-one. Intuitively, I know that the function must be onto, yet I'm having putting it down on paper.
To show the homeomorphism, I know that I have to show that either there is a continuous inverse function of $f$ or show that $f$ is a bijection and that every subset $O$ of $[a,b]$ is open if and only if $f(O)$ is open. To approach it using the second method I know that open sets in $\mathbb{R}$ are open intervals and for continuous functions the image of each interval is an interval, but do we know that an open interval maps to an open interval?
Thank you for any hints on how to approach this.
 A: Since $f$ is increasing, it is injective (proof?). One should actually prove, in my opinion, that $f([a,b])=[f(a),f(b)]$. It is clear that $f([a,b])\subseteq [f(a),f(b)]$ since $f$ is monotone. Can you prove the other inclusion? You do need the intermediate value theorem here! With this out of the way, one has a one one, onto function of the form $$f:[a,b]\to[c,d]$$
It's inverse then exists. Can you show it is continuous?
A: Surjectivity Hints:


*

*Let $y \in \lbrack f(a),f(b) \rbrack$. Let $x_1 = \sup \{x \in \lbrack a,b \rbrack : f(x) < y\}$ and $x_2 = \inf\{x \in \lbrack a,b\rbrack : f(x) > y\}$. What's true about $x_1$ and $x_2$?

*Using the answer you found to my first question and the continuity of $f$, can you prove that $f(x_1) = y$? (The easiest way to do this does not use IVT.)

*This will prove that $\lbrack f(a),f(b) \rbrack \subset f(\lbrack a,b\rbrack)$. What about the opposite containment? (It's easier.)
Continuity Hints:


*

*You now have a bijective function $f$ from $I = \lbrack a,b\rbrack$ onto another interval. Bijective $\implies$ $f^{-1}$ exists.

*Let $\epsilon > 0$ and $a \leq x_0 \leq b$. The continuity of $f$ implies there exists $\delta > 0$ such that for all $x \in \lbrack a,b\rbrack$ such that $|x-x_0| < \delta$ we have $|f(x) - f(x_0)| < \epsilon$. Rewrite these inequalities in terms of $f^{-1}(y)$ and $f^{-1}(y_0)$ where $y = f(x)$ and $y_0 = f(x_0)$. Alternatively, note that this says $f((x_0-\delta,x_0+\delta)) \subset (f(x_0)-\epsilon,f(x_0)+\epsilon)$ and reach a conclusion from that.

A: From increasing you can easily show that $f$ is injective. 
Let's show that $f:[a,b]\rightarrow\left[f(a),f(b)\right]$ is onto. If it's not onto and $S=\{x\in[a,b]:[f(a),f(x)]\subseteq[f(a),f(b)]\}, T=\{x\in[a,b]:[f(x),f(b)]\subseteq[f(a),f(b)]\}$ then by considering the $\sup S$  and $\inf T$ and taking the limits of $f(x)$ as $x\to\sup S$ and $x\to\inf T$ you can derive a contradiction. 
