MA3421 (Functional Analysis 1) Tutorial sheet 1
[September 26, 2014]
Name: Solutions
1. Consider the set N as a submetric space of the real line (with the usual distance on R).
Show that the closed ball B̄(6, 1) = {5, 6, 7} but that the closure of the open ball B(6, 1)
is different.
Solution: In the setting of (N, d) where d is the usual absolute value distance (given by
d(n, m) = |n − m| for n, m ∈ N), the ball B̄(6, 1) of radius 1 about 6 is
B̄(6, 1) = {m ∈ N : d(m, 6) ≤ 1}.
But, for n, m ∈ N, d(m, 6) ≤ 1 means |m − 6| ≤ 1 or −1 ≤ m − 6 ≤ 1 or equivalently
6 − 1 = 5 ≤ m ≤ 6 + 1 = 7. The only integers m with 5 ≤ m ≤ 7 are 5, 6 and 7 and so
B̄(6, 1) = {5, 6, 7}.
Using similar logic the open ball B(6, 1) = {m ∈ N : d(m, 6) < 1} is B(6, 1) = {6}
(singelton set). In a metric space singleton sets are always closed.
(For instance in this case, for n 6= 6 the open ball B(n, d(n, 6)) = B(n, |n − 6|) is in the
complement of {6}, showing that N \ {6} is open. But in fact B(n, 1) = {n} for each
n ∈ N and so all singleton sets are open, hence all subsets of N are open, or N has the
discrete topology. So all subsets are open in this case, and also all subsets are closed since
their complements are open.)
So the closure of the open ball B(6, 1) = {6} is just itself (and not the same as the closed
ball).
What is the boundary of B(6, 1) in this situation?
Solution: For any subset E ⊆ N, ∂E = Ē \ E ◦ . But for E = B(6, 1), E is both open and
closed so that E ◦ = E and (as we just checked) Ē = E also. So ∂E = E \ E = ∅.
2. In the metric space (R2 , d), where d is the standard euclidean metric, show that the maps
π1 : R2 → R and π2 : R2 → R given by π1 (x1 , x2 ) = x1 and π2 (x1 , x2 ) = x2 are continuous.
Solution: Notice that for x = (x1 , x2 ) ∈ R2 and y = (y1 , y2 ) ∈ R2 , we have
p
d(x, y) = (x1 − y1 )2 + (x2 − y2 )2
whereas π1 (x) = x1 , π1 (y) = y1 , and so the distance (in R)
p
p
d(π1 (x), π1 (y)) = |x1 − y1 | = (x1 − y1 )2 ≤ (x1 − y1 )2 + (x2 − y2 )2 = d(x, y)
Now given ε > 0 and a point p ∈ R2 , we can take δ = ε and then we have δ > 0 so that
d(x, y) < δ ⇒ d(π1 (x), π1 (y)) ≤ d(x, y) < δ = ε.
In summary
d(x, y) < δ ⇒ d(π1 (x), π1 (y)) < ε.
For each ε > 0 we can find δ > 0. So π1 is continuous at p. As this is the case for each
p ∈ R2 , π1 is continuous.
The proof that π2 is continuous is exactly similar as
p
p
d(π2 (x), π2 (y)) = |x2 − y2 | = (x2 − y2 )2 ≤ (x1 − y1 )2 + (x2 − y2 )2 = d(x, y)
3. Let X be a set and let T be the collection of all subsets of X that are either the empty set or
cofinite subsets (that is have finite complements). Show that T is a topology on X (which
we call the cofinite topology).
Solution: We need to verify that T satisfies the axioms for a topology on X:
(i) φ ∈ T and X ∈ T ;
We have φ ∈ T by definition of T .
To show X ∈ T note that the complement of X in X is X \ X = ∅, which is finite.
So X is cofinite in X, X ∈ T .
S
(ii) if Ui ∈ T for all i ∈ I = some index set, then i∈I Ui ∈ T ;
So, letSUi ∈ T for all i ∈ I. There are two cases. We could have Ui = ∅ for all i ∈ I.
Then i∈I Ui = ∅ ∈ T . Otherwise there is j ∈ I with Uj 6= ∅. For that Uj we have
X \ Uj finite (as ∅ =
6 Uj ∈ T ). Using de Morgan’s laws
!
\
[
X\
X \ Ui ⊆ X \ Uj
Ui =
i∈I
S
i∈I
Hence X \
i∈I Ui is finite (subset of a finite set is finite) and so
(being cofinite).
S
i∈I
Ui ∈ T
(iii) if U1 , U2 ∈ T , then U1 ∩ U2 ∈ T .
Let U1 , U2 ∈ T . Here there are two cases also. We could have (at least) one of U1 or
U2 the empty set. Then U1 ∩ U2 = ∅ ∈ T . In the remaining case neither U1 nor U2 is
the empty set and then both must have finite complements. Using de Morgan’s laws
X \ (U1 ∩ U2 ) = (X \ U1 ) ∪ (X \ U2 )
is finite (union of two finite sets) and so U1 ∩ U2 is cofinite, thus U1 ∩ U2 ∈ T in this
case also.
Richard M. Timoney
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