Continuous Function Maps Open Interval to Open Sets? Given a continuous function $f$. Then is it true that $\{f(x):x\in (a,b)\}$ is open and $\{f(x):x \in (a,b]\}$ connective? 
I know in general a continuous function doesn't map open sets to open sets if it's not open mapping. But I can't find a counterexample here.
 A: A continuous function $f:\mathbb R \to\mathbb R$ is open if and only if $f$ is strictly monotone.  
Suppose $f$ is not strictly monotone.  Then there exist $x<y<z$ such that $f(y)$ is not strictly between $f(x)$ and $f(z)$; WLOG (because we would consider $-f$) suppose $f(y)\geq f(x)$ and $f(y)\geq f(z)$.  By continuity $f$ attains a maximum value on $[x,z]$, and because of the last sentence this maximum value must be attained on the interior.  Hence $f((x,z))$ has a maximum value, which implies it is not open.
Suppose $f$ is strictly monotone.  WLOG (because we could consider $-f$) suppose $f$ is strictly increasing.  Then for all $x<y$, $f((x,y))=(f(x),f(y))$, using that $f$ is increasing and using the Intermediate Value Theorem.  
So constant functions give counterexamples, but also any function not always increasing or always decreasing, like $x^2$, $\sin(x)$, $\frac{x}{1+x^2}$, or a nowhere differentiable continuous function.
Note that strict monotonicity is equivalent to injectiveness, so $\mathbb R$ has the property that every continuous injective $f:\mathbb R\to \mathbb R$ is an open map, hence is a homeomorphism onto its image.
(Pawel already answered the connectedness part, which could also be thought of in terms of the Intermediate Value Theorem.)
A: Regarding your first question, consider a constant function $f(x)=0$. Then it is a continuous function that maps an open set (open interval) to a set that is not open (point).
Regarding your second question, continuous functions map connected sets to connected sets. Thus, since any interval is connected, its image will also be connected.
