1st December 2004

Munkres §13

Ex. 13.1 (Morten Poulsen). Let (X, T ) be a topological space and A ⊂ X. The following are equivalent:

(i) A ∈ T .
(ii) ∀ x ∈ A ∃ U_{x} ∈ T : x ∈ U_{x} ⊂ A.

Proof. (i) ⇒ (ii): If x ∈ A then x ∈ A ⊂ A and A ∈ T .

(ii) ⇒ (i): A = ⋃_{x∈A} U_{x}, hence A ∈ T . □

Ex. 13.4 (Morten Poulsen). Note that every collection of topologies on a set X is itself a set: A topologi is a subset of P(X), i.e. an element of P(P(X)), hence a collection of topologies is a subset of P(P(X)), i.e. a set.

Let {T_{α}} be a nonempty set of topologies on the set X.

(a). Since every T_{α} is a topology on X it is clear that the intersection ⋂ T_{α} is a topology on X.

The union ⋃ T_{α} is in general not a topology on X: Let X = {a, b, c}. It is straightforward to check that T_{1} = {X, ∅, {a}, {a, b}} and T_{2} = {X, ∅, {c}, {b, c}} are topologies on X. But T_{1} ∪ T_{2} is not a topology on X, since {a, b} ∩ {b, c} = {b} ∉ T_{1} ∪ T_{2}.

(b). The intersection of all topologies that are finer than all T_{α} is clearly the smallest topology containing all T_{α}.

The intersection of all T_{α} is clearly the largest topology that is contained in all T_{α}.

(c). The topology T_{3} = T_{1} ∩ T_{2} = {X, ∅, {a}} is the largest topology on X contained in T_{1} and T_{2}.

The topology T_{4} = {X, ∅, {a}, {b}, {a, b}, {b, c}} is the smallest topology that contains T_{1} and T_{2}.

Ex. 13.5 (Morten Poulsen). Let (X, T ) be a topological space, A basis for T and let {T_{α}} be the set of topologies on X that contains A.

Claim 1. T = ⋂ T_{α}.

Proof. ”⊂”: Let U ∈ T . By lemma 13.1, U is an union of elements of A. Since T_{α} is a topology for all α, it follows that U ∈ T_{α} for all α, i.e. U ∈ ⋂ T_{α}.

”⊃”: Clear since A ⊂ ⋂ T_{α} ⊂ T . □

Now assume A is a subbasis.

Claim 2. T = ⋂ T_{α}.

Proof. ”⊂”: Let U ∈ T . By the definition of a subbasis and the remarks at the bottom on page 82, U is an union of finite intersections of elements of A. Since T_{α} is a topology for all α, it follows that U ∈ T_{α} for all α, i.e. U ∈ ⋂ T_{α}.

”⊃”: Clear since A ⊂ ⋂ T_{α} ⊂ T . □

Ex. 13.6 (Morten Poulsen). The topologies R_{l} and R_{K} on R are not comparable:

R_{l} ⊄ R_{K}: Consider [−1, 0) ∈ R_{l}. Clearly no basis element B_{K} ∈ R_{K} satisfy −1 ∈ B_{K} ⊂ [−1, 0), hence R_{K} is not finer than R_{l}, by lemma 13.3.

R_{K} ⊄ R_{l}: Consider (−1, 1) − K ∈ R_{K}. Clearly no basis element B_{l} ∈ R_{l} satisfy 0 ∈ B_{l} ⊂ (−1, 1) − K, hence R_{l} is not finer than R_{K}, by lemma 13.3.

Ex. 13.7 (Morten Poulsen). We know that T_{1} and T_{2} are bases for topologies on R. Furthermore T_{3} is a topology on R. It is straightforward to check that the last two sets are bases for topologies on R as well.

The following table show the relationship between the given topologies on R.

T_{1}  T_{2}  T_{3}  T_{4}  T_{5}
T_{1} =  ⊂ (1) ⊄ (2) ⊂ (3) ⊄ (4)
T_{2} ⊄ (5) = ⊄ (6) ⊂ (7) ⊄ (8)
T_{3} ⊂ (9) ⊂ (10) = ⊂ (11) ⊄ (12)
T_{4} ⊄ (13) ⊄ (14) ⊄ (15) = ⊄ (16)
T_{5} ⊂ (17) ⊂ (18) ⊄ (19) ⊂ (20) =

(1) Lemma 13.3.

(2) Since R − (0, 1) not finite.

(3) Given basis element (a, b) ∈ T_{1} and x ∈ (a, b) then the basis element (a, x] ∈ T_{4} satisfy x ∈ (a, x] ⊂ (a, b), hence T_{4} is finer than T_{1}, by lemma 13.3.

(4) Given a basis element (a, b) ∈ T_{1} and x ∈ (a, b) then there are clearly no basis element (−∞, c) ∈ T_{5} such that x ∈ (−∞, c) ⊂ (a, b), hence T_{5} is not finer than T_{1}, by lemma 13.3.

(5) Lemma 13.3.

(6) Since R − (0, 1) not finite.

(7) Given basis element (a, b) − K ∈ T_{2} and x ∈ (a, b) − K. If x ∈ (0, 1) then there exists m ∈ Z_{+} such that \frac{1}{m} < x < \frac{1}{m−1}, hence x ∈ (\frac{1}{m}, x] ⊂ (a, b) − K. If x ∉ (0, 1) then x ∈ (a, x] ⊂ (a, b) − K. It follows from (4) and lemma 13.3 that T_{4} is finer than T_{2}.

(8) Since T_{1} ⊄ T_{5} and T_{1} ⊂ T_{2}.

(9) Let U ∈ T_{3}, U nonempty, i.e. R − U = {r_{1}, …, r_{n}}, r_{1} < ⋯ < r_{n}. Since

U = (⋃_{i=1}^{∞}(r_{1} − i, r_{1})) ∪ ( ⋃_{j=1}^{n−1}(r_{j}, r_{j+1})) ∪ ( ⋃_{k=1}^{∞}(r_{n}, r_{n} + k))

it follows that U ∈ T_{1}.

(10) Since T_{3} ⊂ T_{1} ⊂ T_{2}.

(11) Let U ∈ T_{3}, U nonempty, i.e. R − U = {r_{1}, …, r_{n}}, r_{1} < ⋯ < r_{n}, and let x ∈ U. If d = min\{ |x − r_{i}| | i ∈ \{1, …, n\} \} > 0 then x ∈ (x − \frac{d}{2}, x + \frac{d}{2}] ⊂ U. It follows from lemma 13.3 that T_{4} is finer than T_{3}.

(12) Consider U = R − {0} ∈ T_{3}. There are no basis element (−∞, a) ∈ T_{5} such that 1 ∈ (−∞, a) ⊂ U, hence T_{5} is not finer than T_{4}, by lemma 13.3.

(13) Given basis element (c, x] ∈ T_{4} there is clearly no basis element (a, b) ∈ T_{1} such that x ∈ (a, b) ⊂ (c, x], hence T_{1} is not finer than T_{4}, by lemma 13.3.

(14) Given basis element (c, x] ∈ T_{4} there is clearly no basis element B_{K} ∈ T_{2} such that x ∈ B_{K} ⊂ (c, x], hence T_{2} is not finer than T_{4}, by lemma 13.3.

(15) Since R − (0, 1] not finite.

(16) Given basis element (c, x] ∈ T_{4} there is clearly no basis element (−∞, a) ∈ T_{5} such that x ∈ (−∞, a) ⊂ (c, x], hence T_{5} is not finer than T_{4}, by lemma 13.3.

(17) Since (−∞, a) = ⋃_{i=1}^{∞}(a − i, a) ∈ T_{1} for all a ∈ R.

(18) Since T_{5} ⊂ T_{1} ⊂ T_{2}.

(19) Since R − (−∞, 0) not finite.

(20) Given basis element (−∞, a) ∈ T_{5} and x ∈ (−∞, a) then clearly x ∈ (x−|x−a|, x+ \frac{|x−a|}{2}] ⊂ (−∞, a), hence T_{4} is finer than T_{5}, by lemma 13.3.

Ex. 13.8 (Morten Poulsen).

(a). Let

B = { (a, b) | a, b ∈ Q, a < b }.

It is straightforward to check that B is a basis. Let T be the standard topology on R generated by the basis:

{ (r, s) | r, s ∈ R }.

Let U ∈ T and let x ∈ U. Then (by definition of an open set in a topology generated by a basis) there exists a basis element (r, s), r, s ∈ R, such that x ∈ (r, s). Furthermore there exists a, b ∈ Q such that r ≤ a < x < b ≤ s, hence x ∈ (a, b) ⊂ (r, s). It follows, by lemma 13.2, that B is a basis for T .

(b). Let

C = { [a, b) | a, b ∈ Q, a < b }.

It is straightforward to check that C is a basis. Let T_{C} be the topology on R generated by C. Consider [\sqrt{2}, 2) ∈ R_{l}. There are clearly no basis element [a, b) ∈ C such that \sqrt{2} ∈ [a, b) ⊂

[\sqrt{2}, 2), hence T_{C} is not finer than R_{l}, by lemma 13.3.

Since R_{l} is clearly finer than T_{C}, it follows that R_{l} is strictly finer than T_{C}.

References