Strictly monotonic increasing function with a closed domain and range 
Let $a,b,c,d \in \mathbb{R}$ with $a<b$, $I  = [a,b]$. Let $f: I \rightarrow \mathbb{R}$ be a monotonic, strictly increasing function. Also $c<d$ and $f([a,b]) =[c,d]$

a) Proof that $f$ is continuous
b) Proof that the inverse function of $f$ exists and that it's continuous.  
For question a) I already know that f is injective but that is all I can think of. Obviously I have to use the fact that $f$ is monotonic and that the domain and range are closed but still I don't see how to proof this.
For question b it's easy to show that the inverse exists. Because if $f$ is continuous then it surjective, which you proof by using the intermediate value theorem. But then how to show the inverse is also continuous I have no clue
 A: For continuity, fix $\varepsilon>0$. Consider the sets $A=(f(x_0)-\varepsilon/2,f(x_0))$ and $B=(f(x_0),f(x_0)+\varepsilon/2)$. If $x_0\neq a$,there must be a point $x_1\in[a,b]$ such that $f(x_1)\in A$. If $x_0\neq b$, there must be a point $x_2$ such that $f(x_2)\in B$. Since $f$ is strictly increasing, $x_1<x_0<x_2$. Let $\delta=\min\{x_0-x_1,x_2-x_1\}$. Then for all $x$ such that $x_1<x<x_2$, $f(x_1)<f(x)<f(x_2)$, so $$|f(x)-f(x_0)|\leq|f(x)-f(x_1)|+|f(x_1)-f(x_0)|<\varepsilon.$$
If $x_0=a$, we let $\delta=x_2-x_0$, where $x_2$ is as described above, and then the result follows in the same manner.
If $x_0=b$, we let $\delta=x_0-x_1$, where $x_1$ is as described above, and the result follows in the same manner.
If $f:[a,b]\rightarrow [c,d]$ is bijective, monotonic and continuous, $f^{-1}:[c,d]\rightarrow [a,b]$ is bijective and monotonic, so it is continuous for the same reason as $f$.
A: Is $F([A,B])=[C,D]$ is just notation for the fact that the image of $F$ is an interval?
O does it denote that its $\text{co-dom(F)}$
Is equal to identified with its $\text{Im(F)}$ F's image? 
Or that the $\text{co-domain(F)} = [C,D]$, and always equal to its $\text{Im(F)}$ where both are, both intervals?
Otherwise, its unclear that the function $F:[A,B]\to\mathbb{R}$
would be bi-jective. That is ,if the \text{codom(F)}$ is the real number line. 
Unless C,D=-$\infty,\infty$ respectively where $[C,D]=\text{IM(F)}$
I presume that the main part of the question deals with the continuity of $F$ and of inverse in $(2)$. 
Wouldn't this hold even if $F$ monotone increasing.
I presume then that if it does, the inverse function would then exist and be strictly increasing as well due to the in-jectivity of $F$ 
if it $F$ were only monotone, and not in-jective by way of being strictly monotone than the inverse function may not be a traditional many to one function). 
That the inverse exists is a bit to answer unless one knows what the co-domain of $F$ is it the entire real number line ? And in that case what is meant by the inverse function existing?
Is $(A)\mathbb{R}$ the co-domain?, or 
$(B)$ $[C,D]=IM(F)$? 


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Ff the former, $(A)$.
I cant see how $\text{Im(F)}=[c,d]=\text{co-domain(F)}=\mathbb{R}$,
Ie that the $\text{co-dom(F)}$ the entire real number line. 
As would be required for $F$ to be sur-jective (its image to be identical with its co-domain).
 and thus for$F$ to be bi=jective, given that its already in-jective via strict monotony? 
On the other hand, if $(B)$, than even if $F$ were not continuous,  the restriction of the function $F$ to its own image will be always be bi-jective and strictly monotone and so will its inverse , though neither will  necessarily continuous. As the corresponding domains, cod-domains of the restricted inverse functions may not be intervals? 
$$F:[A,B]=\text{dom(F)}\to \text{Im(F)}$$,
As this restriction $\text{IM(F)}$, will generally always be, its sur-jective by definition, for any function, as as $F$ is strictly monotone, injective as well. And thus  bi-jective and strictly monotone for strictly  monotone $F$. Where the inverse over this restricted co-domain.
existent restricted inverse function,though not necessarily continuous, nor will image always be an interval/new co-domain be an interval.
and $\text{dom}(F)_{\text{res\text{IM(F)}}}^{-1}$ the domain,  of the inverse function of the $F_{\text{res-co-dom:Im(F)}}$ restricted to function odomain may not exactly be connected) whether $F$ is continuous or not.
But it in some sense it exists, and is bi-jective, and strictly monotonic,though not necessarily continuous)


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They both imply continuity of $F's$ restriction,  if $F$ is strictly monotonic on closed and bounded, domain with image an closed and bounded interval, as the restricted function will not only be bijective and strictly monotonic (whether continuous or not) defined on a closed domain, but that as the co-domain of the restricted F, (restricted to be $[C,D]$, F's image, is its new co-domain,) then asts image, is also an interval, it is continuous -. 
Bi-jectivity is not enough. if the co-domain is not an interval. Of course the question specifies that the restriction, has a co-domain (or rather the image of $F$) is an interval. 
So i presume that unless the range of co-domain of  a strictly monotonic increasing surjective F is a connected set or an closed and bounded interval, $F$ need not be continuous (as otherwise any strictly monotonic F will have a continuous restriction). 
Which is not the case as far as I know?
OR do these mean the same things (I thought that meant in-tern). 
