1.2. a) A = (0, 1) ∪ (1, 2) is not an open domain, since A = [0, 2], so Int(A) =
(0, 2).
b) Let C be a closed set, and let A = Int(C). Then A ⊂ C, hence A ⊂ C =
C, therefore Int(A) ⊂ Int(C) = A.
On the other hand, A ⊃ A, hence Int(A) ⊃ Int(A) = A.
c) Let us show that for all sets A, B we have Int(A ∩ B) = Int(A) ∩ Int(B).
The set Int(A)∩Int(B) is open and is contained in A∩B, hence Int(A)∩Int(B) ⊂
Int(A ∩ B). On the other hand Int(A ∩ B) is open and is contained in A, hence
Int(A ∩ B) ⊂ Int(A). Same is true for B, so Int(A ∩ B) ⊂ Int(A) ∩ Int(B).
Note that A ∩ B is closed and contains A ∩ B, hence A ∩ B ⊃ A ∩ B for any
sets A and B. It is not true that A ∩ B = A ∩ B in general.
Suppose that A and B are open domains. Then Int(A ∩ B) ⊂ Int(A ∩ B) =
Int(A) ∩ Int(B) = A ∩ B. On the other hand, Int(A ∩ B) ⊃ Int(A ∩ B) = A ∩ B.
1.4. If C is a countable dense subset of a metric space X, then the set of open
balls of the form B1/n (x) for x ∈ C is a basis of topology.
If B = {Ui }i∈I is a countable basis of topology, then choose xi ∈ Ui . Then
{xi }i∈I is dense.
1.5. b) If the Sorgenfrey line is second countable, then its subspace [0, +∞)
is also second countable. Let {Ui }i∈I be a countable basis of topology. Let
A = {inf Ui }i∈I . It is a countable set of real numbers, hence there exists x0 ∈
[0, +∞) \ A. It is easy to see that [x0 , x0 + 1) can not be represented as a union
of elements of {Ui }i∈I .
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