Does a connected space implies continuity and vice versa? I was thinking about this question when we gone over limit and continuity in my complex analysis class.
Let a region $R$ be the image of a function $F$, if I can prove that $R$ is a connected space can I assume that $F$ is continuous at the inverse image of $R$?
Intuitively, I feel that connectivity of an image is a requirement of continuity as otherwise you can find a point $z-z_0$ that satisfies the following
$$|z-z_0|<\delta$$ for which $$|F(z)-F(z_0)|>\epsilon$$
 A: I'm sure you'll agree it is possible for a function $F$ to be discontinuous at a point $p$ in its domain. Suppose that $F$ is discontinuous at some point $p$. Let $R=\{f(p)\}$ be the region that is the image of just this one point. The region $R$ is certainly connected. But $F$ is not continuous on the inverse image of $R$, since it is not continuous at $p$.

A function can be discontinuous and still have a connected image. For example, consider the function $F:\mathbb{R}\to\mathbb{R}$ defined by
$$F(x)=\begin{cases}
x+1 & \text{if }x<0,\\
x & \text{if }x\geq 0.
\end{cases}$$
Its image is all of $\mathbb{R}$, which is connected, but $F$ is not continuous at $x=0$.
Moreover, a function can have a disconnected image and still be continuous. For example, let $S$ be the union $(0,1)\cup (2,3)$, and consider the identity function $F:S\to S$, that is, the function $F(x)=x$. Then $F$ is continuous, but its image (namely, $S$) is not connected.
A: You're right that connectivity of an image is a requirement of continuity.  You're wrong that if $R$ is connected, then $f$ is continuous.  Counterexample: Let $f(x) = 1-x$ if $x < 1$, $f(x) = x$ otherwise.  Then $R = [0, \infty)$, but $f$ is discontinuous at $x=1$.
