If $ f $is continuous on an open interval is the range,$ f((a,b))$, an open interval? Given that $ f $ is a continuous real-valued function on $[a,b]$, then $f([a,b])$ is a closed interval. I am wondering if this fact follows for open intervals. Thanks
 A: If you have a closed (and bounded) interval $[a, b]$, then it is compact and connected. These are both properties which are respected by continuous functions in general, so $f([a, b])$ is compact and connected, and in $\Bbb R$, that means a closed interval.
For an open interval $(a, b)$, you can tell that $f((a, b))$ is connected, so it is an interval, but in general you cannot say what kind of interval (open, closed or half-open). Examples of all three:


*

*Open: $f(x) = x$ for any open interval $(a, b)$.

*Half-open: $f(x) = x^2$ for any open interval $(a, b)$ where $a<0<b$.

*Closed: $f(x) = \sin(x)$ for any open interval $(a, b)$ with $b-a > 2\pi$.

A: It does not, a simple counterexample is the function $f(x)=1$ with $I=(0,1)$. Here $f((0,1))=\{ 1\} $ which is not open in $\mathbb{R}$. Continuity means that the inverse image of open sets are open, hence, given any continuous function $f$ and open set $V$ in its codomain, $f^{-1}(V)$ must be open in its domain.
A: With intermediate value theorem you can easily prove that if $f$ is continuous and injective, then the set $f((a,b))$ is indeed open.
