This is from Hatcher's Algebraic Topology, page 30.

I thought that the circle, S1 is path connected. How then can it be decomposed into the disjoint union of open sets? Furthermore, how can two disjoint open arcs in S1 be the have S1 as their union? What happened to the boundary of the two open sets?

Also note that a disjoint union of open sets A and B does not, in general, require that A and B be disjoint, although, as Tac-Tics has noted, in this case the inverse image of an open arc in S^{1} is the union of disjoint open subsets of R.

But p was defined to be p(s)=(cos 2*pi*s, sin 2*pi*s). Doesn't that mean that the preimage of a cover in S1 must cover an interval in R? Also, it says that each p-1(U_alpha) was mapped homeomorphically to U_alpha by p. Doesn't that mean that connectedness must be preserved by the inverse function?

<quote>Also note that a disjoint union of open sets A and B does not, in general, require that A and B be disjoint</quote>

That makes sense! So that means that the use of disjoint here is redundant since when you take a union, you never choose the same element twice. Am I right?

You're right! That makes so much more sense now! The disjoint open sets would then just be the periodic repeats of the arcs. I need to read more carefully. Thank you so much - you've made my day! :)