A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

124 CHAPTER 16. TOPOLOGY AND COMPLETENESS
16.2.3 The Topology of QN (is Weird)
Definition 16.2.10 (Connected). Let X be a topological space. A subset S of X
is disconnected if there exist open subsets U1, U2 ⊂ X with U1 ∩ U2 ∩ S = ∅ and
S = (S ∩ U1) ∪ (S ∩ U2) with S ∩ U1 and S ∩ U2 nonempty. If S is not disconnected
it is connected.
The topology on QN is induced by dN, so every open set is a union of open balls
B(x, r) = {y ∈ QN : dN(x, y) < r}.
Recall Proposition 16.2.8, which asserts that for all x, y, z,
d(x, z) ≤ max(d(x, y), d(y, z)).
This translates into the following shocking and bizarre lemma:
Lemma 16.2.11. Suppose x ∈ QN and r > 0. If y ∈ QN and dN(x, y) ≥ r, then
B(x, r) ∩ B(y, r) = ∅.
Proof. Suppose z ∈ B(x, r) and z ∈ B(y, r). Then
a contradiction.
r ≤ dN(x, y) ≤ max(dN(x, z), dN(z, y)) < r,
You should draw a picture to illustrates Lemma 16.2.11.
Lemma 16.2.12. The open ball B(x, r) is also closed.
Proof. Suppose y ∈ B(x, r). Then r ≤ d(x, y) so
B(y, d(x, y)) ∩ B(x, r) ⊂ B(y, d(x, y)) ∩ B(x, d(x, y)) = ∅.
Thus the complement of B(x, r) is a union of open balls.
The lemmas imply that QN is totally disconnected, in the following sense.
Proposition 16.2.13. The only connected subsets of QN are the singleton sets {x}
for x ∈ QN and the empty set.
Proof. Suppose S ⊂ QN is a nonempty connected set and x, y are distinct elements
of S. Let r = dN(x, y) > 0. Let U1 = B(x, r) and U2 be the complement of
U1, which is open by Lemma 16.2.12. Then U1 and U2 satisfies the conditions of
Definition 16.2.10, so S is not connected, a contradiction.