Prove that if a function is increasing on [a,b] and satisfied the intermediate value property, then that function is continuous on [a,b]. Let $f$ be increasing on $[a,b]$, i.e. for all $x<y$ in $[a,b]$, $f(x)\leq f(y)$. Also assume $f$ satisfies the intermediate value property. Show that $f$ is continuous on $[a,b]$.
My attempt: Let $f$ be increasing on $[a,b]$ and satisfy IVP, i.e. $\forall x<y$ on $[a,b]$ and $\forall L$ between $f(x)$ and $f(y)$, $\exists c\in(x,y)$ with $f(c)=L$.
Let $\epsilon>0$. Let $c\in(a,b)$. To show $f$ continuous on $[a,b]$, we need to choose a $\delta>0$ such that whenever $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$.
Since $f$ is increasing and $a<c$, then $f(a)\leq f(c)$. I want to show that $f(c)-\epsilon\leq f(x)\leq f(c)+\epsilon$ but don't know how.
 A: Consider a point $c\in[a,b]$ where $f$ is supposed not continous. So $\exists\epsilon_0>0$ such that there is NO $\delta>0$ with the property that: 
if |$x-c$|<$\delta$, then |$f(x)-f(c)$|<$\epsilon_0$.
Now, since $f$ is increasing on [a,b], it is one-to-one on [a,b]. This allows us to look at the expression
$f(c)-\epsilon_0<f(c)<f(c)+\epsilon_0$, and realize that there must be an $x_1,x_2\in[a,b]$ such that $x_1<x_2$ and,
$f(x_1)=f(c)-\epsilon_0$
$f(x_2)=f(c)+\epsilon_0$
So we have $f(x_1)<f(c)<f(x_2)$, and since $f$ is increasing, $x_1<c<x_2$.
So now choose $\delta$=min{$x_2-c,c-x_1$}
Now with this $\delta$ in mind, if |$x-c$|<$\delta$, it must be the case that $f(x_1)=f(c)-\epsilon_0<f(x)<f(c)+\epsilon_0=f(x_2)$.
But we suppose that there wasn't any $\delta$ that would work for this $\epsilon_0$.
A: Let $f^-(a)=f(a)$ and let $f^-(x)=\lim_{x'\to x^-}f(x')=\lim_{x'\to x\land x'<x}f(x')$ for $x\in (a,b].$ 
Similarly $f^+(b)=f(b)$ and let $f^+(x)=\lim_{x'\to x^+}f(x')=\lim_{x'\to x\land x'>x}f(x')$  for $x\in [a,b)$.
$f^-$ and $f^+$ exist because $f$ is monotonic. And because $f$ is increasing we have $f^-(x)\leq f(x)\leq f^+(x)$ for all $x\in [a,b].$ 
Therefore $f$ is continuous iff $f^-(x)=f^+(x)$ for all $x\in [a,b].$
Because $f$ is increasing we have 
(I).  $f^-(x)=\sup_{x'\in [a,x)}f(x')$ when $x\in (a,b].$ 
(II).  $f^+(x)=\inf_{x'\in (x,b]}f(x')$ when $x\in [a,b). $
By contradiction suppose  $f^+(x)-f^-(x)=r>0$ then at least one $y\in  \{f^+(x)-f(x), f(x)-f^-(x)\}$ is positive.  
If $f(x)<y<f^+(x)$  then by the def'n of $f^+(x)$ we must have $x<b$ so by (II) we have  $\forall x'\in (a,b]\; (y<f^+(x)\leq f(x'))$ .  So $x<b$ and $f(x)<y<f(b)$ but no $x'\in  (x,b)$ can satisfy $f(x')=y$.  
Similarly if $f^-(x)<y<f(x)$ then $a<x$ and (I) applies, implying that $f(a)<y<f(x)$ but no $x'\in (a,x)$ satisfies $f(x')=y.$ 
