MA3421 (Functional Analysis 1) Tutorial sheet 2
[October 9, 2014]
Name: Solutions
1. Let X be a set and let T be the collection of all subsets of X that are either the empty set
or cocountable subsets (that is have countable complements). Then T is a topology on X
(which we call the cocountable topology). [You can take that as true, no need to verify it
now.]
Show that if a sequence (xn )∞
n=1 in X converges to a limit ` ∈ X, then there is N so that
xn = ` for each n > N . (In words, the only convergent seqeunces are eventually constant.)
Solution: Suppose (xn )∞
n=1 is a sequence in X with xn → ` as n → ∞ (where ` ∈ X).
Let A = {xn : xn 6= `}, so that A ⊂ X is countable. Then let U = X \ A, an open set in
X as the complement X \ U = A is countable (which means that U is cocountable). Also
` ∈ U (as ` ∈
/ A).
By definition of “xn → ` as n → ∞”, there is n0 so that xn ∈ U for all n > n0 . But
xn ∈ U means xn ∈
/ A and that means xn = `. So xn = ` for all n > n0 .
2. Let X = {0, 1} and S = {{0}}. What is the topology with S as subbase? (Justify.)
[Clarification: S has just one subset of X in it, {0}.]
Solution: Any topology contains ∅ and X. So a topology that contains S certainly must
contain
{∅, X, {0}}
But that is already a topology. So it is the smallest one containing S .
3. Let (XT ) be a toplogical space and B a base fot the topology T . Then let B1 = {B ×C :
B, C ∈ B}. Show that B1 is a base for a topology on X × X. [Corrected version.]
Solution: We just need two properties for B1 to make it a base for some toplology on
X ×X
(a) union of all the sets is the whole space X × X.
The sets B × C are all contained X × X. If x = (x1 , x2 ) ∈ X × X, then there are
B, C ∈ B with x1 ∈ B ⊆ X and x2 ∈ C ⊆ X (since B is a base and X is open, or
because the union of the sets in B must be X). So
x = (x1 , x2 ) ∈ B × C
and so x is in the union of the sets in B1 . As this holds for each x ∈ X,
[
X ×X ⊆
B×C ⊆X ×X
B×C∈B1
and so we have equality (as required).
(b) “intersections of basic open sets open property”
If x = (x1 , x2 ) ∈ (B1 × C1 ) ∩ (B2 × C2 ) with B1 × C1 , B2 × C2 ∈ B1 , then
x1 ∈ B1 ∩ B2 and x2 ∈ C1 ∩ C2 . Since B is a base, there are B3 , C3 ∈ B with
x1 ∈ B3 ⊆ B1 ∩ B2 and x2 ∈ C3 ⊆ C1 ∩ C2 .
Hence we have B3 × C3 ∈ B1 with
x = (x1 , x2 ) ∈ B3 × C3 ⊆ (B1 × C1 ) ∩ (B2 × C2 ).
This is the second property needed for a collection of subsets to be a base.
Hence B1 is a base for some topology on X × X.
Richard M. Timoney
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