= Solution
If $f(A)=U\cup V$ were a <separation>, then
$$
A\cap f^{-1}(U),\qquad A\cap f^{-1}(V)
$$
would be disjoint nonempty relatively <open sets> covering $A$. They are relatively <open> by <continuity> of the restriction $f|_A:A\to f(A)$. This contradicts <connectedness> of $A$.
$$
\boxed{A\text{ connected},\ f\text{ continuous}\ \Longrightarrow\ f(A)\text{ connected}.}
$$
This is the <continuous image of a connected space> theorem; no separation property of the target is required.
Back to article page
