Proof that a continuous function maps connected sets into connected sets I'm trying to prove that, if f is a function from C to C, and its domain, D, is connected, then f(D) is also connected. How would I go about doing this?
The definition of conectedness at play is "S is disconnected iff there exist open disjoint sets A and B such that none contains S, but their union does", and that of continuity is "f is continuous iff, if a sequence of members of the domain tends to z, then the image of the sequence tends to f(z)".

Suppose that $g:f(D)\rightarrow\{0,1\}$ is continues, (where we equip $\{ 0, 1 \}$ with the discrete topology $\big\{ \emptyset, \{ 0 \}, \{ 1 \}, \{ 0, 1 \} \big\}$) then $g \circ f$ is continues. Since $D$ is connected, we deduce that $g\circ f$ is constant. This implies that $g$ is constant since  the image of the restriction of $f$ to $D$ is $f(D)$. We deduce that $f(D)$ is connected.
