Show that the image of an open interval under a continuous, strictly monotone function is an open interval. I have written a proof. In addition to feedback, I would appreciate alternative proofs.
note: need to update for non-finite sup or inf.
Exercise: Show that the image of an open
interval under a continuous, strictly monotone function is an open interval.
Proof:
let $f:(a,b) \to I$. assume that $f$ is continuous and strictly monotone.
assume $f$ strictly increasing
let $g:[a,b] \to J$. where $g(x)$=\begin{cases}f(x) & x\in(a,b)\\ s=\sup\{f(x):x\in(a,b)\} & x=b \\ i=\inf\{f(x):x\in(a,b)\} & x=a \\\end{cases}
claim: $g(x)$ is continuous at $b$.
let $\epsilon > 0$.
now, $g(b)-\epsilon=s-\epsilon$ is not an upper bound for $f$.
so, there exists $y \in (a,b)$ such that $f(y)=g(y)>g(b)-\epsilon$
let $\delta=b-y$
now, $|b-x|<\delta$ implies $x>y$
now if $x< b $, because $f$ is strictly increasing we have: $g(x)>g(y)$
so, $g(b)-\epsilon<g(x)\le s<g(b)+\epsilon$ implies $|g(x)-g(b)|< \epsilon$.
so, $g$ is continuous at $b$. a similar argument shows  $g$ is continuous at $a$
this, combined with the fact that $f$ is continuous on $(a,b)$, implies that $g$ is continuous on its entire domain.
let $y \in (g(a),g(b))$.
by I.V.T. there exists $c \in [a,b]$ such that $g(c)=y$
note $c\not =b$ and $c \not = a $.
so there exists $c \in (a,b)$ such that $g(c)=f(c)=y$
for all $c \in (a,b)$, $f(c) \not = s$ and $f(c) \not = i$
because if there was such a $c$, then pick $x \in (c,b)$
we have $x>c$ and so $f(x)>f(c)$, contradicting $s$ being the supremum.
I have shown that there exists $c∈(a,b)$ such that $f(c)=y$ for all $y∈(g(a),g(b))$ and, for all $y∈\mathbb{R} \setminus (g(a),g(b))$, there is no $c$ such that $f(c)=y$. Therefore, the image of $f$ is an open interval: $(g(a),g(b))$.
 A: Your approach is fine, but it has a little gap: It is possible that $i = -\infty$ or $s = \infty$. But these cases are even easier because you do not have to consider boundary points $i, s \in \mathbb R$.
A: Your task is to prove two things:

*

*The image of $f$ is an interval.

*This interval is open.

What you have proved explicitly is only this:


*The image of $f$ has no maximum or minimum.

But that's not that far off the mark. To complete your proof, you would need to argue that firstly, the image is indeed an interval, and secondly that an interval without maximum and minimum is open.
To prove that the image is an interval you can use the IVT.
And another point: the image of the function might not have a supremum or infimum, and then your argument breaks. But the good thing is, you don't actually need that. After all, you're trying to prove that the image doesn't have a maximum or minimum. Which is obviously true if it doesn't even have a supremum or infimum. But you should clearly state this.
