ASSIGNMENT 4
TOPOLOGY
CLOSED SETS.
IGOR WIGMAN
PLEASE INFORM ME IF THERE ARE ANY INQUIRIES/MISPRINTS.
(1) (a) By the definition ∂A = A \ Int(A). For a point x to be in A it is necessary and
sufficient that every neighbourhood of x intersects A, and x ∈
/ Int(A) if and only if
every neighbourhood U intersects X \ A.
(b) If x ∈ A, then, in particular, x ∈ A, so only the condition x ∈
/ Int(A) remains to
be checked.
(c) We saw that always A = A ∪ A0 , the question being when the union is disjoint.
Suppose A∩A0 = ∅ and let x ∈ A. If x is not isolated, then for every neighbourhood
U of x there exists a point y ∈ U ∩ A, and then x ∈ A0 ∩ A, which is a contradiction
to A ∩ A0 = 0.
Conversely, suppose that A consists of isolated points only and assume by contradiction, that there exists a point x ∈ A ∩ A0 . Since x ∈ A0 , every neighbourhood of
x intersects A \ {x}, which is a contradiction since x ∈ A is isolated.
(d) By the definition,
\
A=
C.
A⊆C closed
Since any C closed relatively to T is also closed relatively to T 0 , for the latter the
intersection is over a larger (in the sense of inclusion) family of sets resulting in a
smaller intersection, i.e. AT 0 ⊆ AT . Similarly, IntT (A) ⊆ IntT 0 (A).
(2) (a)
A = A, Int(A) = ∅, ∂A = A, A0 = A.
(b)
B 0 = B = {(x, y) : x ≥ 0}, Int(B) = B
∂B = {(x, y) : y = 0, x ≥ 0} ∪ {(x, y) : x = 0}.
(c)
C 0 = C = B ∪ {(x, y) : y = 0}
∂C = {(x, y) : (y = 0 and x ≤ 0) or x = 0}, Int(C) = {x > 0}.
(d)
D0 = D = ∂D = R2 , Int(D) = ∅
(3) (a) A = R \ Q the set of irrational numbers inside X = R, and inside Y = R \ Z.
ĀR = A0R = R, Int(A)R = ∅, ∂AR = R.
∂AY = A0Y = ĀY = ĀR ∩ Y = R \ Z, A0R = R, Int(A)Y = ∅.
(b) Appeared in class (partially).
(c) Note that the topology in this set is finer than the standard one, so that the corresponding inclusions follows for the closure and the interior from Q1(d).
The tricky part in this question is to determine which of the endpoints belong
to the closure/interior/limit set. We choose to work with part (2) of the natural
characterization of the closure, which states that to check whether x belongs to
the closure of a set, it is enough to check whether its every basic neighbourhood
intersects A.
In all the situations, (α, β) is contained in A0 , A and Int(A); (R \ [α, β]) ∩ A = ∅,
(R \ [α, β]) ∩ A0 = ∅, {α, β} ⊇ ∂A. Also, necessarily α ∈ A and α ∈ A0 , regardless
1
2
ASSIGNMENT 4 SOLUTIONS
on whether it is in A or not, since any neighbouhood [α, α + ) contains points in
A other than α. There is no symmetry here so β may not be in A.
To finish with x = α, we already mentioned that α ∈ A, A0 ; α ∈ Int(A) iff α ∈ A,
since then [α, α + ) ⊆ A is some neighbourhood of α; α ∈ ∂A, iff α ∈
/ A.
Now let us deal with x = β. In this case β ∈
/ A0 , since [β, β + ) will not contain
points of A other than (possibly) β itself. However, β ∈ A iff β ∈ A. On the other
hand, β ∈
/ Int(A) always and then β ∈ ∂A, iff β ∈ A.
All in all
(
(α, β) α ∈
/A
Int(A) =
[α, β) α ∈ A,
(
[α, β) β ∈
/A
A=
[α, β] β ∈ A,
A0 = [α, β)

∅



{α}
∂A =

{β}



{α, β}
α ∈ A, β ∈
/A
α, β ∈
/A
.
α, β ∈ A
α∈
/ A, β ∈ A
(4) (a) Let x ∈ A ∪ B i.e. every neighbourhood of x intersects A ∪ B. Assume by contradiction that x ∈
/ A and x ∈
/ B. It means that there exist neighbourhoods U1 , U2
of x not intersecting A or B respectively. Then U := U1 ∩ U2 is a neighbourhood
of x that doesn’t intersect A ∪ B, which is a contradiction. Note that here we use
the finiteness of the union in that intersection of only finitely many neighbourhoods
(open sets containing x) is a neighbourhood. The other inclusion is true in general
(see the next
S solution).
Ai . Then x ∈ Ai0 for some i0 ∈ I. Then every neighbourhood of x
(b) Let x ∈
i∈I
S
S
Ai so that x ∈
Ai indeed.
intersects Ai0 , and in particular, it intersects
i∈I
i∈I
To see a counter-example for the equality in the infinite case take e.g. Ai = {ri }i∈N
where {ri } is some ordering of the set of rational numbers Q (which is countable).
(c) The mistake is that that the index i depends on the neighbourhood U , so one cannot
infer that
T
T x ∈ Ai for any fixed i.
(d) If x ∈
Ai , then for every U neighbourhood of x, U intersects
Ai , i.e. for every
i∈I
i∈I
T
i, U intersects Ai . Hence for every i, x ∈ Ai i.e. x ∈
Ai .
i∈I
Note that the converse inclusion is not true (e.g. A1 = (0, 1) and A2 = (1, 2)); the
reason a similar argument to the above would not work is that if a neighbourhood U
of some point intersects all of Ai , it is not guaranteed to intersect their intersection
(as it may intersect the various sets in different points).
(5) (a) Under the discrete topology a sequence {xn } ⊆ X is convergent, only if xn stabilizes
for n sufficiently big, i.e. there exists an N s.t. for all n ≥ N , xn = x for some
x ∈ X; in this case xn → x.
(b) For the first task it is sufficient to notice that if for every neighbourhood U of x in
T , xn ∈ U , then the same holds for neighbourhoods of x in T 0 , since as T 0 is coarser
than T , any such neighbourhood is in particular in T .
To construct such an example (one of many) take xn = n1 ∈ R at first equipped with
the standard topology, and at second with the finite complement topology Tf . For
the latter topology xn → x for every number x ∈ R, much like the same holds for
the example yn = n considered in class (though lacking the convergence property in
R). Note that the same (that is a sequence converging to any number x ∈ R) holds
in Tf for any sequence not attaining a value infinitely many times. You may want
ASSIGNMENT 4
TOPOLOGY
CLOSED SETS.
3
to think what happens if this condition violated (such as, for example, in the other
considered example xn = (−1)n ).
n
(6) (a) Take for example (−1)
n . To see that it is not convergent in Rl (to 0) choose the
basic neighbourhood [0, 0.1) (say).
(b) A sequence {xn } is convergent to x in Rl if it is convergent to x in R and, in addition,
xn ≥ x for n sufficiently large.