Download Topology 440, Homework no. 2 Solutions

February 6, 2006
Topology 440, Homework no. 2
(Due on Wednesday, February 22)
1. Let X be a topological space and A a subset of X. Show that
X − A = X − Int(A)
and
Int(X − A) = X − A
Use these equalities to show that ∂A = ∂(X − A).
2. Let (X, <) be a simply ordered set and let Tσ be the order topology on X. Show
that with respect to the order topology the inclusion a, b ⊆ [a, b] holds for any two
elements a, b ∈ X with a < b. Under what conditions is the equality a, b = [a, b]
true?
3. For subsets A ⊆ X and B ⊆ Y of the topological spaces X and Y , show that
A × B = A × B in the product topology on X × Y .
4. Show that X is a Hausdorff space if and only if the diagonal Δ = {(x, x) | x ∈ X} is
a closed set in X × X (equipped with the product topology).
March 1, 2006
Solutions
1. (a) X − A = X − Int(A)
This follows from the definitions of the closure and interior of the set A. Let B
and C be the collections of subsets of X defined by
B = {B ⊆ X | X − A ⊆ B, B is a closed set }
C = {C ⊆ X | C ⊆ A, C is an open set }
Observe that there is a bijective correspondence between these two collection,
the one which associates to a set C ∈ C to the set B = X − C ∈ B. On the other
hand
X −A=
B
and
Int(A) =
C
B∈B
C∈C
Therefore, using DeMorgan’s laws, we have
C=
(X − C) =
B = X −A
X − Int(A) = X −
C∈C
C∈C
B∈B
2
(b) Int(X − A) = X − A
Follows by the same principle as part (a). Let D and E be the collections
D = {D ⊆ X | D ⊆ X − A, D is an open set }
E = {E ⊆ X | A ⊆ E, E is a closed set }
These are again bijective sets under the function D → X − E. As in part (a) a
simple use of DeMorgan’s laws completes the job:
E=
(X − E) =
D = Int(X − A)
X −A=X −
E∈E
E∈E
D∈D
(c) This part is an exercise in set theory:
∂(X − A) = X − A − Int(X − A)
= (X − Int(A)) − (X − A)
= A − Int(A)
The fact that (X −Int(A))−(X −A) = A−Int(A) can easily be checked directly
but becomes obvious from the picture below.
X − Int(A)
X −A
A − Int(A)
Figure 1. The big oval in each of the 3 pictures is the set X. The dotted line
indicates the set Int(A) and the polygon encloses A. The set A itself is not visible.
The shaded region in each picture represents the set listed under the diagram. The
difference of the first two sets clearly equals the last set.
2. Since [a, b] = −∞, a∪b, ∞, the latter of which is an open set in the order topology,
it follows that [a, b] is a closed set. On the other hand a, b ⊆ [a, b] and therefore
a, b ⊆ [a, b] (because the closure is the smallest closed set containing a, b).
For the other part of the question, notice that [a, b] = a, b ∪ {a, b}. Thus the only
possibilities for a, b are
⎧
[a, b] = a, b ∪ {a, b}
⎪
⎪
⎨
[a, b = a, b ∪ {a}
a, b =
a, b] = a, b ∪ {b}
⎪
⎪
⎩ a, b
Each of the 4 cases above can occur. If a, b = [a, b then [a, b is a closed set and so
X − [a, b = −∞, a ∪ [b, ∞ is an open set. This for example happens if there is an
3
element c ∈ X, c < b such that there is no element x ∈ X with c < x < b. In that
case [b, ∞ = c, ∞.
Similarly if there is an element d ∈ X, a < d such that there is no x ∈ X with
a < x < d then a, b] is a closed set. If both c and d exist then a, b is itself a closed
set.
3. This was shown in class.
4. =⇒ Suppose X is Hausdorff. We will show that Δ is closed in X × X by showing
that X ×X −Δ is an open set. Pick an arbitrary point (a, b) ∈ X ×X −Δ. Then a = b
(since otherwise (a, b) ∈ Δ) and so the Hausdorff property guarantees the existence
of open sets Ua , Vb with a ∈ Ua , b ∈ Vb and Ua ∩ Vb = ∅. These properties in turn
imply that (a, b) ∈ Ua × Vb and Ua × Vb ⊂ X × X − Δ. Since (a, b) was arbitrary we
can write
X ×X −Δ=
Ua × Vb
(a,b)∈X×X−Δ
The sets on the right-hand side are open sets in the product topology.
⇐= Assume that Δ is closed in X × X. To show that X is Hausdorff, pick two
arbitrary points a, b ∈ X (a = b). We need to find to open, disjoint sets U, V with
a ∈ U, b ∈ V . Since Δ is closed, X × X − Δ is open and is therefore a union of open
sets:
X × X − Δ = ∪i Ui × Vi
where Ui , Vi is some family of open subsets of X. Find the index j (there may be
many such indices but there is for sure at least one!) for which (a, b) ∈ Uj × Vj . Then
a ∈ Uj and b ∈ Vj . If c ∈ Uj ∩ Vj then (c, c) ∈ (Uj × Vj ) ∩ Δ which is impossible since
Uj × Vj ⊆ X × X − Δ. Thus Uj ∩ Vj = ∅ as needed.