If X is not connected, and U,V are two disjoint non empty open subsets that cover X, let p be in U and q in V. Then continuously map an interval f:I-->X so that f(0) = p and f(1) = q. Then f^-1(U) and f^-1(V) are disjoint non empty open sets covering I, a contradiction since I is connected.
The point is that I is connected and the image of a connected set is connected, hence any path connected set is connected.
Just out of curiosity, how would you prove to this class that the plane itself is connected without this result?
thinking again about this topic, the hard part of course is to prove the unit interval is connected in the sense of open sets. then after that, proving other sets are connected seems to proiceed naturally via the route outlined here, i.e. iamges of connected sets are conncted and then path connected implies connected. the theorem that products of connected sets are connected seems more difficult, maybe easier to prove products preserve path connectedness.
If you knew star shaped sets are connected, then one could write the twice punctured plane as a union of two such with overlap, but how to prove that?
in gheneral the easiest way to prove facts about sets you know are connected is to use the fact that a set is connected iff every continuous map from it to a 2 point set is constant, but that does not easily help greatly to establish connectivity of a specific set like an interval. or rather, evben with that aid, you still need to use the lub property.
