This was a shower thought.

• We usually define a topological space $X$ as connected iff there are disjoint open sets $U, V$ such that $U \cup V = X$. Since they are disjoint, we have that $U \cap V = \emptyset$.
• An alternative way of stating this is to consider two colors C = {red, blue } with the discrete topology.
• We use the discrete topology on C since we want the two colors to be "separate".
• Now, a space $X$ is connected iff there is a continuous surjective function $f: X \rightarrow C$. That is, we can color the whole space continuously with both colors.

This is equivalent to the original definition by setting $U = f^{-1}(red)$ and $V = f^{-1}(blue)$:

• Pre-images of a function must be disjoint. Hence, $U \cap V = \emptyset$.
• Preimages of $red$ and $blue$ must be open sets since $\{red\}$ and $\{blue\}$are open and $f$ is continuous: continuous functions have pre-images of open sets as open. Hence $U$ and $V$ are open.
• Since $f$ is surjective, we must have that the pre-images cover the entire set $X$. Hence $U \cup V = X$.

I find this to be appealing, since it's intuitively obvious to me that if a space is disconnected, I can color it continuously with two colors, while if a space is connected, I should be unable to color it continuously with two colors --- there should be a point of "breakage" where we suddenly switch colors.