1st December 2004

Munkres §17

Ex. 17.3. A × B is closed because its complement

(X × Y) − (A × B) = (X − A) × Y ∪ X × (Y − B)

is open in the product topology.

Ex. 17.6.

(a). If A ⊂ B, then all limit points of A are also limit points of B, so [Thm 17.6] \overline{A} ⊂ \overline{B}.

(b). Since A ∪ B ⊂ \overline{A} ∪ \overline{B} and \overline{A} ∪ \overline{B} is closed [Thm 17.1], we have \overline{A ∪ B} ⊂ \overline{A} ∪ \overline{B} by (a). Conversely, since A ⊂ A ∪ B ⊂ \overline{A ∪ B}, we have \overline{A} ⊂ \overline{A ∪ B} by (a) again. Similarly, \overline{B} ⊂ \overline{A ∪ B}. Therefore \overline{A} ∪ \overline{B} ⊂ \overline{A ∪ B}. This shows that closure commutes with finite unions.

(c). Since ⋃ A_{α} ⊃ A_{α} we have \overline{⋃ A_{α}} ⊃ \overline{A_{α}} by (a) for all α and therefore \overline{⋃ A_{α}} ⊃ ⋃ \overline{A_{α}}. In general we do not have equality as the example A_{q} = {q}, q ∈ Q, in R shows.

Ex. 17.8.

(a). By [Ex 17.6.(a)], \overline{A ∩ B} ⊂ \overline{A} and \overline{A ∩ B} ⊂ \overline{B}, so \overline{A ∩ B} ⊂ \overline{A} ∩ \overline{B}. It is not true in general that \overline{A ∩ B} = \overline{A} ∩ \overline{B} as the example A = [0, 1), B = [1, 2] in R shows. (However, if A is open and D is dense then \overline{A ∩ D} = \overline{A}).

(b). Since ⋂ A_{α} ⊂ A_{α} we have \overline{⋂ A_{α}} ⊂ \overline{A_{α}} for all α and therefore \overline{⋂ A_{α}} ⊂ ⋂ \overline{A_{α}}. (In fact, (a) is a special case of (b)).

(c). Let x ∈ \overline{A} − \overline{B}. For any neighborhood of x, U − \overline{B} is also a neighborhood of x so

U ∩ (A − B) = (U − B) ∩ A ⊃ (U − \overline{B}) ∩ A ≠ ∅

since x is in the closure of A [Thm 17.5]. So x ∈ \overline{A − B}. This shows that \overline{A}−\overline{B} ⊂ \overline{A − B}. Equality does not hold in general as \overline{R} − \overline{\{0\}} = R − \{0\} ⫋ \overline{R − \{0\}} = R.

Just to recap we have

(1) A ⊂ B ⇒ \overline{A} ⊂ \overline{B} (A ⊂ B, B closed ⇒ \overline{A} ⊂ B)

(2) \overline{A ∪ B} = \overline{A} ∪ \overline{B}

(3) \overline{A ∩ B} ⊂ \overline{A} ∩ \overline{B} (\overline{A ∩ D} = \overline{A} if D is dense.)

(4) \overline{⋃ A_{α}} ⊃ ⋃ \overline{A_{α}}

(5) \overline{⋂ A_{α}} ⊂ ⋂ \overline{A_{α}}

(6) \overline{A} − \overline{B} ⊂ \overline{A − B} Dually,

(1) A ⊂ B ⇒ Int A ⊂ Int B (A ⊂ B, A open ⇒ A ⊂ Int B)

(2) Int (A ∩ B) = Int A ∩ Int B
(3) Int (A ∪ B) ⊃ Int A ∪ Int B

These formulas are really the same because

\overline{X − A} = X − Int A, Int (X − A) = X − \overline{A}

Ex. 17.9. [Thm 19.5] Since \overline{A} × \overline{B} is closed [Ex 17.3] and contains A × B, it also contains the closure of A × B [Ex 17.6.(a)], i.e. \overline{A × B} ⊂ \overline{A} × \overline{B}.

Conversely, let (x, y) ∈ \overline{A} × \overline{B}. Any neighborhood of (x, y) contains a product neighborhood of the form U × V where U ⊂ X is a neighborhood of x and V ⊂ Y a neighborhood of y. The intersection of this product neighborhood with A × B

(U × V) ∩ (A × B) = (U ∩ A) × (V ∩ Y)

is nonempty because U ∩ A ≠ ∅ as x ∈ \overline{A} and V ∩ B ≠ ∅ as y ∈ \overline{B}. Since thus any neighborhood of (x, y) intersect A × B nontrivially, the point (x, y) lies in the closure of A × B [Thm 17.5]. This shows that \overline{A} × \overline{B} ⊂ \overline{A × B}.

Ex. 17.10 (Morten Poulsen).

Theorem 1. Every order topology is Hausdorff.

Proof. Let (X, ≤) be a simply ordered set. Let X be equipped with the order topology induced by the simple order. Furthermore let a and b be two distinct points in X, may assume that a < b. Let

A = { x ∈ X | a < x < b },

i.e. the set of elements between a and b.

If A is empty then a ∈ (−∞, b), b ∈ (a, ∞) and (−∞, b) ∩ (a, ∞) = ∅, hence X is Hausdorff. If A is nonempty then a ∈ (−∞, x), b ∈ (x, ∞) and (−∞, x) ∩ (x, ∞) = ∅ for any element x in

A, hence X is Hausdorff. □

Ex. 17.11 (Morten Poulsen).

Theorem 2. The product of two Hausdorff spaces is Hausdorff.

Proof. Let X and Y be Hausdorff spaces and let a_{1} × b_{1} and a_{2} × b_{2} be two distinct points in X × Y. Note that either a_{1} ≠ a_{2} or b_{1} ≠ b_{2}.

If a_{1} ≠ a_{2} then, since X is Hausdorff, there exists open sets U_{1} and U_{2} in X such that a_{1} ∈ U_{1}, a_{2} ∈ U_{2} and U_{1} ∩ U_{2} = ∅. It follows that U_{1} × Y and U_{2} × Y are open in X × Y. Furthermore a_{1} × b_{1} ∈ U_{1} × Y, a_{2} × b_{2} ∈ U_{2} × Y and (U_{1} × Y) ∩ (U_{2} × Y) = (U_{1} ∩ U_{2}) × Y = ∅ × Y = ∅, hence X × Y is Hausdorff.

The case b_{1} ≠ b_{2} is similar. □

Ex. 17.12 (Morten Poulsen).

Theorem 3. Every subspace of a Hausdorff space is Hausdorff.

Proof. Let A be a subspace of a Haussdorff space X and let a and b be two distinct points in A.

Since X is Hausdorff there exists two open sets U and V in X such that a ∈ U, b ∈ V and U ∩ V = ∅. Hence a ∈ A ∩ U, b ∈ A ∩ V and (A ∩ U) ∩ (A ∩ V) = (U ∩ V) ∩ A = ∅ ∩ A = ∅. Since A ∩ U and A ∩ V are open in A, it follows that A is Hausdorff. □

Ex. 17.13 (Morten Poulsen).

Theorem 4. A topological space X is Hausdorff if only if the diagonal

Δ = { x × x ∈ X × X | x ∈ X }

is closed in X × X.

Proof. Suppose X is Hausdorff. The diagonal Δ is closed if and only if the complement Δ^{c} = X × X − Δ is open. Let a × b ∈ Δ^{c}, i.e. a and b are distinct points in X. Since X is Hausdorff there exists open sets U and V in X such that a ∈ U, b ∈ V and U ∩ V = ∅. Hence a × b ∈ U × V and U × V open in X × X. Furthermore (U × V) ∩ Δ = ∅, since U and V are disjoint. So for every point a × b ∈ Δ^{c} there exists an open set U_{a×b} such that a × b ∈ U_{a×b} ⊂ Δ^{c}. By Ex. 13.1 it follows that Δ^{c} open, i.e. Δ closed.

Now suppose Δ is closed. If a and b are two distinct points in X then a × b ∈ Δ^{c}. Since Δ^{c} is open there exists a basis element U × V, U and V open in X, for the product topology, such that a × b ∈ U × V ⊂ Δ^{c}. Since U × V ⊂ Δ^{c} it follows that U ∩ V = ∅. Hence U and V are open sets such that a ∈ U, b ∈ V and U ∩ V = ∅, i.e. X is Hausdorff. □

Ex. 17.14 (Morten Poulsen). The sequence converges to every real number, by the following result.

Theorem 5. Let X be a set equipped with the finite complement topology. If (x_{n})_{n∈Z_{+}} is an infinite sequence of distinct points in X then (x_{n}) converges to every x in X.

Proof. Let U be a neighborhood of x ∈ X, i.e X − U is finite. It follows that x_{n} ∈ U, for all, but finitely many, n ∈ Z_{+}, i.e. (x_{n}) converges to x. □

Ex. 17.21 (Morten Poulsen). Let X be a topological space. Consider the three operations on P(X), namely closure A ↦ \overline{A}, complement A ↦ X − A and interior A ↦ A^{∘}. Write A^{−} instead of \overline{A} and A^{c} instead of X − A, e.g. X − \overline{X − A} = A^{c−c}.

Lemma 6. If A ⊂ X then A^{∘} = A^{c−c}.

Proof. A^{∘} ⊃ A^{c−c}: Since A^{c} ⊂ A^{c−}, A^{c−c} ⊂ A and A^{c−c} is open.

A^{∘} ⊂ A^{c−c}: Since A^{∘} ⊂ A, A^{c} ⊂ A^{∘c} and A^{∘c} is closed, it follows that A^{c−} ⊂ A^{∘c}, hence A^{∘} ⊂ A^{c−c}. □

This lemma shows that the interior operation can be expressed in terms of the closure and complement operations.

(a). The following theorem, also known as Kuratowski’s Closure-Complement Problem, was first proved by Kuratowski in 1922.

Theorem 7. Let X be a topological space and A ⊂ X. Then at most 14 distinct sets can be derived from A by repeated application of closure and complementation.

Proof. Let A_{1} = A and set B_{1} = A_{1}^{c}. Define A_{2n} = A_{2n−1}^{−} and A_{2n+1} = A_{2n}^{c} for n ∈ Z_{+}. Define B_{2n} = B_{2n−1}^{−} and B_{2n+1} = B_{2n}^{c} for n ∈ Z_{+}.

Note that every set obtainable from A by repeatedly applying the closure and complement operations is clearly one of the sets A_{n} or B_{n}.

Now A_{7} = A_{4}^{c−c} = A_{4}^{∘} = A_{3}^{−∘}. Since A_{3} = A_{1}^{−c} it follows that A_{3} is open, hence A_{3} ⊂ A_{7} ⊂ A_{3}^{−}, so A_{7}^{−} = A_{3}^{−}, i.e. A_{8} = A_{4}, hence A_{n+4} = A_{n} for n ≥ 4. Similarly B_{n+4} = B_{n} for n ≥ 4.

Thus every A_{n} or B_{n} is equal to one of the 14 sets A_{1}, …, A_{7}, B_{1}, …, B_{7}, this proves the result. □

(b). An example:

A = ((−∞, −1) − {−2}) ∪ ([−1, 1] ∩ Q) ∪ {2}.

The 14 different sets:

\begin{aligned}A_{1} & = ((−∞, −1) − \{−2\}) ∪ ([−1, 1] ∩ Q) ∪ \{2\} \\ A_{2} & = (−∞, 1] ∪ \{2\} \\ A_{3} & = (1, ∞) − \{2\} \\ A_{4} & = [1, ∞) \\ A_{5} & = (−∞, 1) \\ A_{6} & = (−∞, 1] \\ A_{7} & = (1, ∞) \\ B_{1} & = \{−2\} ∪ ([−1, 1] − Q) ∪ ((1, ∞) − \{2\}) \\ B_{2} & = \{−2\} ∪ [−1, ∞) \\ B_{3} & = (−∞, −1) − \{−2\} \\ B_{4} & = (−∞, −1] \\ B_{5} & = (−1, ∞) \\ B_{6} & = [−1, ∞) \\ B_{7} & = (−∞, −1). \end{aligned}

Another example:

A = { 1/n | n ∈ Z_{+} } ∪ (2, 3) ∪ (3, 4) ∪ {9/2} ∪ [5, 6] ∪ { x | x ∈ Q, 7 ≤ x < 8 }.

References