§ Connectedness in terms of continuity

This was a shower thought. This is equivalent to the original definition by setting U=f1(red)U = f^{-1}(red) and V=f1(blue)V = f^{-1}(blue):
  • Pre-images of a function must be disjoint. Hence, UV=U \cap V = \emptyset.
  • Preimages of redred and blueblue must be open sets since {red}\{red\} and {blue}\{blue\}are open and ff is continuous: continuous functions have pre-images of open sets as open. Hence UU and VV are open.
  • Since ff is surjective, we must have that the pre-images cover the entire set XX. Hence UV=XU \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.