27th January 2005

Munkres §23

Ex. 23.1. Any separation X = U ∪ V of (X, T ) is also a separation of (X, T′). This means that

(X, T ) is disconnected ⇒ (X, T′) is disconnected

or, equivalently,

(X, T′) is connected ⇒ (X, T ) is disconnected

when T′ ⊃ T .

Ex. 23.2. Using induction and [1, Thm 23.3] we see that A(n) = A_{1} ∪ ⋯ ∪ A_{n} is connected for all n ≥ 1. Since the spaces A(n) have a point in common, namely any point of A_{1}, their union ⋃ A(n) = ⋃ A_{n} is connected by [1, Thm 23.3] again.

Ex. 23.3. Let A∪⋃ A_{α} = C ∪D be a separation. The connected space A is [Lemma 23.2] entirely contained in C or D, let’s say that A ⊂ C. Similarly, for each α, the connected [1, Thm 23.3] space A ∪ A_{α} is contained entirely in C or D. Sine it does have something in common with C, namely A, it is entirely contained in C. We conclude that A∪⋃ A_{α} = C and D = ∅, contradicting the assumption that C ∪ D is a separation

Ex. 23.4 (Morten Poulsen). Suppose ∅ ⊊ A ⊊ X is open and closed. Since A is open it follows that X − A is finite. Since A is closed it follows that X − A open, hence X − (X − A) = A is finite. Now X = A ∪ (X − A) is finite, contradicting that X is infinite. Thus X and ∅ are the only subsets of X that are both open and closed, hence X is connected.

Ex. 23.5. Q is totally disconnected [1, Example 4, p. 149]. R_{ℓ} is totally disconnected for R_{ℓ} = (−∞, b) ∪ [b, +∞) for any real number b. Any well-ordered set X is totally disconnected in the order topology for

X = (−∞, α + 1) ∪ (α, +∞) = (−∞, α] ∪ [α + 1, +∞)

for any α ∈ X and if A ⊂ X contains α < β then α ∈ (−∞, α + 1) and β ∈ (α, +∞).

Ex. 23.6. X = Int(A) ∪ Bd(A) ∪ Int(X − A) is a partition of X for any subset A ⊂ X [1, Ex 17.19]. If the subspace C ⊂ X intersects both A and X − A but not Bd(A), then C intersects A − Bd(A) = Int(A) and (X − A) − Bd(X − A) = Int(X − A) and

C = (C ∩ Int(A)) ∪ (C ∩ Int(X − A))

is a separation of C.

Ex. 23.7. R = (−∞, r) ∪ [r, +∞) is a separation of R_{ℓ} for any real number r. It follows [1, Lemma 23.1] that any subspace of R_{ℓ} containing more than one point is disconnected: R_{ℓ} is totally disconnected.

Ex. 23.11. Let X = C ∪ D be a separation of X. Since fibres are connected, p^{−1}(p(x)) ⊂ C for any x ∈ C and p^{−1}(p(x)) ⊂ D for any x ∈ D [1, Lemma 23.2]. Thus C and D are saturated open disjoint subspaces of X and therefore p(C) and p(D) are open disjoint subspace of Y. In other words, Y = p(C) ∪ p(D) is a separation.

Ex. 23.12. Assume that the subspace Y is connected. Let X − Y = A ∪ B be a separation of X − Y and Y ∪ A = C ∪ D a separation of Y ∪ A. Then [1, Lemma 23.1]

\overline{A} ⊂ X − B, \overline{B} ⊂ X − A, \overline{C} ⊂ X − D, \overline{D} ⊂ X − C

and

Y ∪ A ∪ B = X = B ∪ C ∪ D

are partitions [1, §3] of X.

The connected subspace Y is entirely contained in either C or D [1, Lemma 23.2]; let’s say that Y ⊂ C. Then D = C ∪ D − C ⊂ Y ∪ A − Y ⊂ A and \overline{D} ⊂ \overline{A} ⊂ X − B. From

\overline{B ∪ C} =^{[Ex17.6.(b)]} \overline{B} ∪ \overline{C} ⊂ (X − A) ∪ (X − D) = (B ∪ Y) ∪ (B ∪ C) ⊂ B ∪ C

\overline{D} ⊂ (X − B) ∩ (X − C) = D

we conclude that B ∪ C = \overline{B ∪ C} and D = \overline{D} are closed subspaces. Thus X = (B ∪ C) ∪ D is a separation of X.

References

[1] James R. Munkres, Topology. Second edition, Prentice-Hall Inc., Englewood Cliffs, N.J., 2000. MR 57 #4063