If $f:[a,b]\rightarrow [f(a),f(b)]$ is increasing and surjective, prove that it is continuous.
Fix $c \in (a,b)$. Take $\epsilon >0$. We then wish to find the set of $x$ such that $|f(x)-f(c)|<\epsilon$ by definition of continuity.
That's as far as I got and I don't know how to proceed. Any suggestions/proofs?
Sincé $f$ is increasing, $f(x^+)$ and $f(x^-)$ both exist. Furthermore
$\tag 1f(x^-)\le f(x)\le f(x^+).$
Now, suppose there is a point $c\in [a,b]$ at which $f$ is not continuous. Then, there are two possibilities:
$a). f(x^-)=f(x^+)$ in which case $f(c)\neq f(x^+)$ because $f$ is assumed discontinuous at $c$. But this contradicts $1).$
$b). f(x^-)<f(x^+),$ in which case at least one of $[f(x^-),f(c)]$ or $[f(c), f(x^+)]$ is a non-degenerate interval, which contradicts the surjectivity of $f$.
Well, let's try to do this using just the basic, strict, definition of $(\epsilon,\delta)-$continuity.
Recall that surjectivity means that for all $z\in [f(a),f(b)]$, there exists a $x\in [a,b]$ such that $f(x)=z$ and that being increasing (not strictly) means that if $x\leq y$, then $f(x)\leq f(y)$.
Towards a contradiction, for it to be non continuous in a point $c$, you need an $\epsilon >0$ such that $\forall\delta>0, \exists y\in]c-\delta, c+\delta[$ such that $\mid f(y)-f(c)\mid \geq \epsilon$.
But then, if there exists such an $\epsilon$ and such a $y$, it would means that in the interval $[f(y),f(c)]$ (wlog $f(y)<f(c)$, note $y\neq c$), is of length at least $\epsilon$, so there exists values inside and by surjectivity, those values have an preimage : $\forall z\in \; ]f(y),f(c)[, \; \exists x\in[a,b]$ such that $f(x)=z$.
But now, $f(a)\leq f(y)< f(x)< f(c) \leq f(b)$ implies by monotonicity (being increasing here) that $x\in\;]y,c[\; \subset\; ]c-\delta,c+\delta [$. This does the magic and comes from the fact that if $A\implies B$, then $\neg(B) \implies \neg(A)$.
So, since this is true $\forall\delta$ and $\forall z$, we can pick $\delta=\epsilon$ for example and $z=f(c)-\frac{\epsilon}{2}$, we have that there exists an element $x$ in $]y,c[$ such that $f(x)=f(c)-\frac{\epsilon}{2}>f(y)$, so $\mid f(x)- f(c) \mid = \mid f(c)-\frac{\epsilon}{2}-f(c)\mid=\frac{\epsilon}{2}<\epsilon$, which means that for $\delta = \epsilon/2$, then the value $y'\in ]c-\frac{\epsilon}{3},c+\frac{\epsilon}{3}[$ so that $\mid f(c)-f(y')\mid \geq \epsilon$ (which has to exists since we assumed non-continuity) can not be in $]c-\frac{\epsilon}{3},c[$ since we have monotonicity implying $f(x)\leq f(y')\leq f(c)$ if it were there and $f(x)$ being at distance $\epsilon/2<\epsilon$ from $f(c)$ and so, $y' \in ]c,c+\frac{\epsilon}{3}[$ is such that $\mid f(y')-f(c) \mid \geq \epsilon$ implies that we have a non empty interval of length $\epsilon$ there were we can pick the value $f(c)<f(c)+\frac{\epsilon}{4}<f(y')$ and there exists again a $x'\in ]c,c+\frac{\epsilon}{3}[$ such that $f(x')=f(c)+\frac{\epsilon}{4}$, so $\mid f(x')-f(c)\mid = \frac{\epsilon}{4}$.
But then it means that we have found an interval around $c, [x,x']$ such that $f(x)-f(x') = f(c)-\frac{\epsilon}{2}-f(c)-\frac{\epsilon}{4} = \frac{3}{4}\epsilon < \epsilon$ so every possible value $x\leq z\leq x'$ is such that $f(c)-\frac{\epsilon}{2}<f(z)<f(c)+\frac{\epsilon}{2}$ which means that for $\delta=\min\{\mid x'-c\mid,\mid x-c\mid\}$ there are no $y'' \in \big[c-\delta, c+\delta\big]$ such that $\mid f(y'')-f(c)\mid \geq \epsilon$ which is a blatant contradiction of our assumption that it wasn't continuous.
PS : this is not "enough" in the sense that I do not talk about the extrema, but you can do it there using the same method. And you may need to be more careful here and there when taking open/closed intervals.
Take $c\in (a,b)$, and let $\epsilon>0$ be given; assume for now that $f(c)+\epsilon\leq f(b)$ and $f(c)-\epsilon\geq f(a)$. Then by surjectivity there exist $x_1,x_2\in[a,b]$ such that $f(x_1)=f(c)-\epsilon$ and $f(x_2)=f(c)+\epsilon$. Set $\delta=\min\{c-x_1,x_2-c\}$, and use the fact that $f$ is increasing.
If you can't shrink $\epsilon>0$ so that $f(c)+\epsilon\leq f(b)$ and $f(c)-\epsilon\geq f(a)$, then $f(c)=f(b)$ and/or $f(c)=f(a)$, so you will have to deal with these cases. Can you finish it from here?
Here is a unique perspective on the problem: Let $[a,b] \subset [0,1]$. This splits $[0,1]$ into two more intervals, $[0,a)$ and $(b, 1]$. Note that
$$f^{-1}[0,a) \cup f^{-1}[a,b] \cup f^{-1}(b,1] = C$$
and as $f$ is surjective, $[a,b]$ actually exists in the image of $f$.
Also, because $f$ is increasing,
$$\inf f^{-1}[0,a) \leq \inf f^{-1}[a,b] \leq \inf f^{-1}(b,1]$$ and
$$\sup f^{-1}[0,a) \leq \sup f^{-1}[a,b] \leq \sup f^{-1}(b,1]$$
Thus as $C$ is an interval, it has to be $f^{-1}[a,b]$ which contains every value between $\inf f^{-1}[a,b]$ and $\sup f^{-1}[a,b]$, so $f^{-1}[a,b]$ is an interval.
The important bit is that $f^{-1}[a,b]$ is an interval. Specifically, it is a closed interval, as $\inf f^{-1}[a,b] = f^{-1}(a) \in f^{-1}[a,b]$ and $\sup f^{-1}[a,b] = f^{-1}(b) \in f^{-1}[a,b]$.
Thus, the pre-image of $f$ maps closed intervals to closed intervals $([a,b] \mapsto [f^{-1}(a), f^{-1}(b)])$, and by extension, closed sets to closed sets. So, $f$ is continuous.
Let $f:I = [a, b] \to J = [c, d]$ be surjective and non-decreasing and let $O_J \subset J$ be an open interval in $J$.
Note that intervals $[a, x), (x, b], [c, y), (y, d]$ which are half-open in the real line are open in the (subspace) topologies of the closed intervals $I, J$.
Firstly show that the pre-image $X = f^{-1}(O_J) \subset I$ is an interval, and secondly that it's open.
Then since open intervals are a topological basis for $J$ it follows that $f$ is continuous.....
Here's a simpler approach. Fix $\epsilon > 0, x_0 \in [a,b]$, and see that there must exist some (at least one) $x' \in [a, b]$ such that $|f(x') - f(x_0)| = \epsilon$ by surjectivity of $f$. Then see that for $\delta = |x' - x_0|$, any $x $ that is $\delta$-close to $x_0$ will have $|f(x) - f(x_0)|$ by monotonicity. You can graph a nondecreasing function to see this, by observing that, for instance, no $x > x_0$ that is $\delta$-close to $x_0$ can be further from $fx_0$ than $fx'$ is, since otherwise the function's graph would have to decrease from $fx$ to $fx'$, a contradiction. A symmetric argument applies to any hypothetical $x < x_0$. The requirement for $[a, b]$ and $[fa, fb]$ to be closed is just so that such $x_0, x'$ are guaranteed to exist to begin with.
It is trivial from the very definition (at least when one assumes strict monotonicity). It is optional to rely on other results. The actual statement is that if the image of a strictly increasing function defined on an open interval is still an interval, then the function is continuous.
Recall: Let $f : I \subseteq \mathbb{R} \rightarrow J \subseteq \mathbb{R}$ be a function defined on an open set $I$, strictly increasing on $I$ and surjective onto $J$ with $J$ an interval of $\mathbb{R}$. By definition, $f$ is continuous at $x_0 \in I$ if, and only if, $$ \forall \varepsilon > 0 \; \exists \delta_1, \delta_2 > 0 \, :\, \forall x \in I \left[ x_0 - \delta_1 < x < x_0 + \delta_2 \; \Longrightarrow \; f (x_0) - \varepsilon < f (x) < f (x_0) + \varepsilon \right] . $$
Proof: It is sufficient to "reduce" $f (x_0) - \varepsilon < f (x) < f (x_0) + \varepsilon$. By taking the inverse function of $f$ (which exists because by assumption $f$ is strictly increasing), we get
$$f^{- 1} (f (x_0) - \varepsilon) < x < f^{- 1} (f (x_0) + \varepsilon) $$ and, therefore it is sufficient to set $\delta_1= x_0 - f^{- 1} (f (x_0) - \varepsilon)$ and $\delta_2 = f^{- 1} (f (x_0) + \varepsilon) - x_0$.
Note: The surjectivity of $f$ is needed so that the inverse images of the values $f (x_0) \pm \varepsilon$ are not empty (at least when $\varepsilon$ is sufficiently small so that $f (x_0) \pm \varepsilon$ are in $J$).
