Proving the Circle's Connectedness Using Projection and Open Covers

[PhysicalAnomaly]

We shall prove (c) using just one special property of the projection p :R -> S1 ,
namely:
„
There is an open cover {U_alpha} of S1 such that for each alpha, p^(-1) („U…_alpha) can be
decomposed as a disjoint union of open sets each of which is mapped homeomorphically
onto U_alpha by p.

For example, we could take the cover {U_alpha} to consist of any two open arcs in S1
whose union is S1 .

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?

Thanks.

 

[Tac-Tics]

I thought that the circle, S1 is path connected.

It is. Therefore, it's connected.

How then can it be decomposed into the disjoint union of open sets?

It can't. Reread the snippet carefully. You take a cover of S1, then run it backwards through p. The result is two disjoint open sets in R (not S1!).

You might think "well, p is continuous and connectedness is a topological property?" But that doesn't matter because p^-1 is not continuous.

 

[George Jones]

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 S¹ is the union of disjoint open subsets of R.

 

[PhysicalAnomaly]

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?

 

[George Jones]

Also, it says that each p-1(U_alpha) was mapped homeomorphically to U_alpha by p.

No, it doesn't say this.

 

[PhysicalAnomaly]

I'm probably interpreting this wrongly but doesn't it say:

disjoint union of open sets each of which is mapped homeomorphically
onto U_alpha by p.

Thanks for being so patient.

 

[George Jones]

"each of which" refers to the individual sets whose union is being taken.

If

$$p^{-1} \left[ U_\alpha \right] = A_1 \cup A_2 \cup ...,$$

then each $A_i$ is mapped homeomorphically onto $U_\alpha$.

 

[PhysicalAnomaly]

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! :)