Download Math 440, Spring 2012, Solution to HW 1 (1) Page 83, 1. Let X be a

Math 440, Spring 2012, Solution to HW 1
(1) Page 83, 1. Let X be a topological space; let A ⊂ X. Suppose that for each x ∈ A there is an open set
U containing x such that U ⊂ A. Show that A is open in X.
Suppose x ∈ A. From the given hypothesis, we have an open set, say Ux such that x ∈ Ux ⊂ A. It is
[
easy to see that A =
Ux :
x∈A
As x0 ∈ A =⇒ x0 ∈ Ux0 ⊂
[
Ux , we have A ⊂ U.
x∈A
Conversely, since each Ux ⊂ A, we have
[
Ux ⊂ A.
x∈A
Since each Ux is open and arbitrary unions of open sets are open, A is an open set.
(2) Page 83, 3. Let X be a set; let Tc be the collection of all subsets U f X such that X − U is either
countable or all of X. We will show that Tc is a topology on X.
First of all, note that if X is a countable set, then Tc is the Discrete topology on X, as every subset
has a countable complement. So this example is more interesting in the case of uncountably infinite X.
(i) Since X − φ = X, by definition, φ ∈ Tc and X − X = φ, which has zero elements, so φ ∈ Tc .
(ii) Finite intersections: If U1 , U2 are nonempty elements of Tc , then X − Ui is countable for each i.
By de Morgan’s laws, we have X − (U1 ∩ U2 ) = (X − U1 ) ∪ (X − U2 ) and finite union of countable sets
is countable. So U1 ∩ U2 ∈ Tc . If either of them is empty, then the intersection is empty, which is also in
Tc .
(iii) Arbitrary unions: Let Ui be elements of Tc ,for i ∈ I. If for some i0 , Ui0 is nonempty, then X − Ui0
is countable. By de Morgan’s laws, we have
[
\
X − ( Ui ) =
(X − Ui ) ⊂ X − Ui0 .
i∈I
As the last set is countable, it follows that
i∈I
[
Ui inTc . Otherwise, if each Ui is empty, then the union is
i∈I
empty as well and therefore it is in Tc .
Part II.
T∞ = { U | X − U is infinite or φ or X }.
Is this a topology on X?
Note that if X is finite, then this defines the Indiscrete topology on X. However, in general (for an infinite set X), this is not a topology. For example, let X = N, U = { 2n | n ∈ N } and V = { 2n+1 | n ∈ N }.
Then U, V ∈ T∞ , but U ∪ V 6∈ T∞ , since X − (U ∪ V ) = {1}, which is neither X, φ or infinite.
(3) Page 83, 8. (a) Apply Lemma 13.2 to show that countable collection
B = { (a, b) | a < b, a, b ∈ Q }.
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is a basis that generates the standard topology on R.
Statement of Lemma 13.2: Let X be a topological space. Suppose that C is a collection of open sets of
X such that for each open set U of X and each x in U , there is an element C of C such that x ∈ C ⊂ U.
Then C is a basis for the topology of X.
Note that the standard topology on R is generated by open intervals (a, b) where a, b ∈ R. So by
definition of a topology generated by a basis, for any open set U and any x ∈ U , we have x ∈ (a, b) ⊂ U,
for some a, b ∈ R. Now, by density of Q in R, there exist c, d ∈ Q such that a ≤ c < x and x < d ≤ b, or,
in other words:
x ∈ (c, d) ⊂ (a, b) ⊂ U and c, d ∈ Q.
By Lemma 13.2, the above B is a basis for the standard topology on R.
(b) Show that the collection C = { [a, b) | a < b, a, b ∈ Q } is a basis that generates a topology different
from the lower limit topology on R.
First we need to check that C satisfies the conditions to be a basis.
(i) For each x ∈ R, there are rationals a, b such that a ≤ x < b.. So there is a member [a, b) of C such
that x ∈ [a, b).
(ii) Condition (ii) follows from the stronger statement that if two members of C have a nontrivial intersection, then the intersection is also a member of C.
[a, b) ∩ [c, d) = [max{a, c}, min{b, d}).
So C is a basis.
Note that the lower limit topology is generated by intervals [x, y) whre x, y ∈ R. So, [π, 10) is an open
set the lower limit topology, and clearly π ∈ [π, 10).
However, since for no rational numbers a, b we have π ∈ [a, b) ⊂ [π, 10), by the definition of the topology
generated by a basis, [π, 10) is not an open set in the topology generated by the basis C.
(4) Page 83, 5. Show that if A is a basis for a topology on X, then the topology, say T , generated by A
equals the intersection of all topologies that contain A. Prove the same if A is a subbasis.
First of all, note that, T is a member of the collection { Tα }, where each Tα is a topology containing
T
A. Therefore Tα ⊂ T . (Note that this part of the argument works even if A is a subbasis.)
Next, recall from Lemma 13.1 that sets in T are unions of sets in A. Since each of the Tα contains A
as a subcollection and being a topology each Tα is closed under taking arbitrary unions, it follows that
T
every member of T is also a member of each Tα . Therefore T ⊂ Tα .
T
Thus, T = Tα .
If A is a subbasis, from the definition of a subbasis we know that sets in T are unions of finite
intersections of sets in A. Since each of the Tα contains A as a subcollection and being a topology each
Tα is closed under taking finite intersections as well arbitrary unions, it follows that every member of T
T
is also a member of each Tα . Therefore T ⊂ Tα .
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