Download Topology Homework 3

Topology Homework 3
Section 3.1 - Section 3.3
Samuel Otten
3.1
(1)
Proposition. The intersection of finitely many open sets is open and the union of finitely many
closed sets is closed.
Proof. Note that S1 ∩ S2 ∩ S3 ∩ · · · ∩ Sn = (((S1 ∩ S2 ) ∩ S3 ) · · · ∩ Sn ) for any family of sets {Si },
i ∈ N, and any natural number n. Thus, for an intersection of finitely many open sets we can take
the intersection pairwise and, by Definition 1(ii), each set along the way is open and the end result
is open.
Similarly, S1 ∪ S2 ∪ S3 ∪ · · · ∪ Sn = (((S1 ∪ S2 ) ∪ S3 ) · · · ∪ Sn ). Thus, for a union of finitely
many closed sets we can take the union pairwise and, by Proposition 1(b), each set along the way
is closed and the end result is closed. ¤
(6)
Proposition. Let F ⊂ N be closed if F contains a finite number of positive integers or F = N.
Then the set F of all closed subsets of N thus defined forms a topology on N. Moreover, this topology
is not induced by any metric.
Proof. We shall employ Proposition 2 of the text to demonstrate that (N, F) is a topological
space. First, by definition, F = N is closed and φ is closed because it contains a finite number of
positive integers, namely zero. Second, let F1 and F2 be elements of F. If at least one of F1 and
F2 is equal to all of N, then F1 ∪ F2 = N which is closed. If this is not the case, then both contain
a finite number of positive integers. Thus, their union contains a finite number of positive integers
and is closed by definition. Third, we consider the intersection of a family of elements of F. The
intersection of sets containing finitely many elements must itself contain finitely many elements.
The worst that the intersection could do is result in N itself, but this is closed. Therefore, F defines
a topology on N.
Suppose a metric D on N did induce (N, F). Since (N, D) is a metric space, it must satisfy
Proposition 12 of Chapter 2, which says that for any closed subset of the space and any point in
its complement, there exist disjoint open subsets such that the closed subset is contained in one
and the point is contained in the other. Consider F = {1, 2, 3} and x = 4. It is obviously the case
that F is closed and x ∈ N − F , so Proposition 12 must hold. Let U be an open set containing
F and let V be an open set containing x such that U ∩ V = φ. Since U and V are disjoint, both
cannot contain infinitely many elements. Therefore, at least one of them must contain finitely many
elements, which leads us to two cases. Suppose U is finite. Then U C = N − U has infinitely many
elements. Moreover, U C =
6 N because it lacks the elements 1, 2, and 3. Hence, U C is not closed.
(Note that an open set in this topology is defined as the complement to some closed set from F.)
Therefore, U is not open, a contradiction. Similarly, if V is finite, then V C is infinite but lacks the
element 3. So, in this case, V C is not closed and we also have a contradiction. This implies that
this topology is not induced by any metric. ¤
1
(7)
Proposition. Define a family F of subsets of R2 as follows: Let F ∈ F if and only if F = R2 .
F = φ, or F is a set consisting of finitely many points together with the union of finitely many
straight lines. Then the topology determined by F is the coarsest in which lines and points are
closed sets.
Proof. Suppose (R2 , F 0 ) is a topological space such that F 0 is strictly coarser than F. This means
F 0 ⊂ F, so there must exist some F ∗ ∈ F such that F ∗ ∈
/ F 0 . By definition, F ∗ consists of finitely
∗
many points and lines. But since F is not an element of F 0 , the points and lines from F ∗ are not
closed sets in F 0 . If they were individually closed, then their unions taken pairwise would have to
be in F 0 by Proposition 1(b). But this union is precisely F ∗ , which we just said was not in F 0 .
Therefore, any topology that is coarser than F does not have all lines and points as closed sets. ¤
It is not possible to have a topology on the real plane in which every line was a closed set,
but every one-point subset was not. This fact follows from Proposition 1(b) which states that the
union of any two closed sets is closed. Since a one-point subset of the real plane corresponds to
an intersection of lines, and lines are closed, the singleton set would also have to be closed. It is
possible, however, to have a topology on the real plane in which every one-point set was closed
but every line was not. This also follows from Proposition 1(b) which implies that the union of
uncountably many closed sets may not be closed, and a line is the union of uncountably many
(closed) one-point sets.
3.2
(5) Example 6 of the text made the claim that
B1 = {N (x, q) | q ∈ Q, x ∈ X} and
B2 = {N (x, 1/n) | x ∈ X and n a positive integer}
are both bases for the topology induced on X = R2 by the Pythagorean metric. We will verify this
claim using our in-class definition of a basis.
For B1 , let x be any element in X. It is clear that x ∈ N (x, q) for any rational q > 0 because
a point is always in its own neighborhood. Choose x ∈ B1 ∩ B2 where B1 , B2 ∈ B1 . This means
that x is in two overlapping neighborhoods with rational radii. Set p equal to the lesser of the
Pythagorean distances from x to the frontiers of the neighborhoods. If p is rational, then we have
found a B3 ∈ B1 such that x ∈ B3 . If p is irrational, then we set q equal to a rational number that
is less than p, which exists because the rationals are dense. Thus, B1 is a basis for the real plane
with the Pythagorean metric.
For B2 , let x be any element in X. Again, x is guaranteed to be in its own neighborhood for
any radius 1/n. Choose x ∈ B1 ∩ B2 where B1 , B2 ∈ B2 . As above, set p equal to the lesser of the
Pythagorean distances from x to the frontiers of the neighborhoods surrounding it. If p = 1/n for
some positive integer n then we are done. If not, then we choose n so that 1/n < p, which exists
because the sequence 1/n converges to zero. Thus, x ∈ B3 ∈ B2 and B2 is also a basis.
(6) We will list all the other open sets in the smallest topologies on N for which the following are
collections of open sets.
• N, φ are the only open sets. (trivial topology)
• N, {1, 2}, {3, 4, 5} and {1, 2, 3, 4, 5}, φ.
• N, {1, 2}, {3, 4, 5}, {1, 4, 7} and {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5}, {3, 4, 5, 7}, {1, 2, 4, 7}, {1}, {4}, φ.
2
3.3
(4)
Proposition. For each x ∈ R2 , let Nx be the set of interiors of all triangles which contain x in
the interior. Then the collection Nx forms a neighborhood system for a topology on R2 . Moreover,
this topology is the same as the topology induced by the Pythagorean metric.
Proof. We will use our in-class definition to show that this is a neighborhood system.
(i) There are obviously triangle interiors that contain x for any point x ∈ R2 . Thus, Nx 6= φ.
(ii) By definition, x is in the interior of the triangle and so x ∈ N for each N ∈ Nx .
(iii) Let N1 and N2 be two sets from Nx . These sets both necessarily contain x. Since the
interior of triangles (excluding the boundary) are open sets in the real plane, and the intersection
of N1 and N2 is not empty, we know that infinitely many points belong to N1 ∩ N2 . Choosing any
three non-collinear points from this set will result in N3 ∈ Nx .
(iv) Let N be any set from Nx and let y be any point from N . Consider the points that lie
halfway between y and each of the three vertices of the triangle at the frontier of N . These three
points are non-collinear, and if we let N 0 be the interior of the triangle formed by the three points
then N 0 contains y. Thus, N 0 ∈ Ny and N 0 ⊂ N . ¤
It is clear to see that this open neighborhood system determines the same topology as that
induced by the Pythagorean metric because circles (D-neighborhoods) can be circumscribed about
any triangular interior and a triangle can be described about any circle.
(6)
Proposition. Let (X, D) be a metric space and {an } a sequence of positive real numbers which
converges to 0. For each x ∈ X, set Nx = {N (x, an ) | n ∈ N}. Then the collection of Nx thus
formed constitutes an open neighborhood system for the topology induced by D.
Proof. (i) Since {an } comprises positive real numbers, let p = a1 > 0. Then N (x, p) ∈ Nx for
each x ∈ X, and Nx 6= φ.
(ii) Since D(x, x) = 0 and an > 0 for each n, we know that x ∈ N (x, an ) for any n. Thus,
x ∈ N for any N ∈ Nx .
(iii) Suppose N1 and N2 are in Nx . Then N1 = N (x, aj ) and N2 = N (x, ak ) for some j, k ∈ N.
Without loss of generality, suppose aj ≥ ak . Then N1 ∩ N2 = N (x, ak ). Since an → 0, we can find
an am that is arbitrarily close to 0. In this case, we choose am < ak and define N3 = N (x, am ).
Hence, N3 ∈ Nx and N3 ⊂ N1 ∩ N2 .
(iv) Let N = N (x, ap ) be a set from Nx and let y be an element in N . If p = ap − D(x, y) (note
that 0 < p ≤ ap ), we know by the convergence of {an } that there exists an aq such that aq < p.
Therefore, N (y, aq ) ∈ Ny is contained in N (x, ap ).
(v) A subset U of X is clearly open if and only if for each x ∈ U there is N ∈ Nx such that
N ⊂ U . This follows because open sets in the topology induced by D are defined in terms of
neighborhoods, and it is neighborhoods that make up Nx . Also, since an → 0, we will always be
able to find N ⊂ U for any x ∈ U .
Therefore, Nx satisfies the textbook definition of an open neighborhood system. ¤
(7) CLAIM: If Nx is a neighborhood system of a set X with the property that each member of
each Nx is finite, then the topology τ on X determined by Nx is the discrete topology.
This claim is false. We shall consider the following counterexample. Let X = {1, 2, 3} and let Nx
have only one element for x = 1, 2, 3; specifically, N1 = {{1, 2}}, N2 = {{2}}, and N3 = {{2, 3}}.
Clearly, Nx is never the empty set and x ∈ N for each Nx (there is only one). It is true that
N3 ∈ Nx always exists such that N3 ⊂ N1 ∩ N2 because, in fact, N1 = N2 = N3 for each Nx .
Moreover, the only element y of an N ∈ Nx that is distinct from x is 2 (in both N1 and N3 ).
But N 0 ∈ Ny is the set containing 2, which is obviously a subset of N in either case. Therefore,
Nx is a neighborhood system by definition. Also, each member is finite. However, Nx does not
determine the discrete topology because {1}, {3}, and {1, 3} are subsets of X that are not the
union or intersection of sets within Nx .
3
3.G
(1) Let B be a basis for a topology τ on X.
a) We will show that τ is the intersection of all topologies containing B. Let T ∈ τ . Then,
by the definition of a basis (p. 44), T is the union of members of B. Also, by part (iii) of the
definition of a topology (p. 40), any topology containing B must also contain the union of any
family of its members. This means T is in any topology containing B and T is in the intersection
of all topologies containing B. Thus, τ is a subset of that intersection. Conversely, let T be any
set from the intersection of all topologies containing B. Since τ is a topology containing B, T ∈ τ .
Thus, the intersection of all topologies containing B is a subset of τ . Therefore, we can conclude
that they are, in fact, equal.
b) The conclusion from part (a) also holds when B is a subbasis. The argument is the same,
except now we use the definition of subbasis (p. 46) which includes the intersection of finitely many
members, and part (ii) of the definition of a basis which also deals with the intersection.
(2) For the set of integers Z, let B be the set of all bi-infinite arithmetic sequences. Note that an
arithmetic sequence is of the form an + b for n ∈ N and some fixed a, b ∈ Z.
a) We will show that B is a basis for a topology τ on Z. First, observe that {. . . , −2, −1, 0, 1, 2, . . . }
is an element of B. Thus, for any x ∈ Z there is a B ∈ B such that x ∈ B. Second, let B1 and B2 be
arbitrary sets from B and let x ∈ B1 ∩ B2 . We know that B1 and B2 contain elements of the form
an + b and cn + d, respectively. Since x belongs to the intersection, x = an1 + b and x = cn2 + d
for some n1 , n2 ∈ N. But x + ac = a(n1 + c) + b = c(n2 + a) + d, so x + ac ∈ B1 ∩ B2 . By adding x
and all the multiples of ac, we will have generated B3 ∈ B such that B3 ⊂ B1 ∩ B2 . Therefore, B
is a basis by definition.
b) Let us consider some simple closed sets in this topology. By definition, the complement
of a closed set is open; in particular, the complement of a closed set is a bi-infinite arithmetic
sequence. So {. . . , −3, −1, 1, 3, . . . } is closed because its complement is the bi-infinite arithmetic
sequence {. . . , −2, 0, 2, . . . }. (Note that both sets just mentioned are simultaneously open and
closed.) Similarly, {. . . , −5, −4, −2, −1, 1, 2, 4, 5, . . . } is closed because its complement is the set of
multiples of 3. Furthermore, there are more complicated closed sets whose complements are some
union of bi-infinite arithmetic sequences, as we shall see below.
c)
Proposition. There are infinitely many prime numbers.
Proof. Let τ be the topology from part (a). Define m to be the set of multiples of m ∈ N. Thus,
2 = {. . . , −4, −2, 0, 2, 4, . . . }
3 = {. . . , −6, −3, 0, 3, 6, . . . }
5 = {. . . , −10, −5, 0, 5, 10, . . . }
..
.
p = {. . . , −2p, −p, 0, p, 2p, . . . }.
The sets p for p a prime number are open because they are bi-infinite arithmetic sequences. However,
it is important to note that the sets p are also closed. Consider the complement of 2, which
is {. . . , −3, −1, 1, 3, . . . }. This is the set of all numbers congruent to 1(mod 2). Similarly, the
complement of 3 is {. . . , −2, −1, 1, 2, . . . }, which is the union of the set of all numbers congruent
to 1(mod 3) and the set of all numbers congruent to 2(mod 3). Since both of these complements
are open, 2 and 3 themselves are closed.
This reasoning can be generalized to p. The complement of p is the union of open sets; namely,
the sets containing all numbers congruent to 1(mod p), 2(mod p),. . . , p − 1(mod p). Since this
union is open, p is closed.
4
Let us observe one more fact before moving on to the heart of the argument. By definition,
a prime number is not a multiple of any natural number preceding it. That is, p ∈
/ p0 for any p0
preceding p. Moreover, if q is a composite number, then q ∈ p∗ for some p∗ preceding q.
Now, suppose to the contrary that the number of prime numbers is finite. Let p be the last
prime. Then 2 ∪ 3 ∪ · · · ∪ p is the union of finitely many closed sets. We proved in Problem (1) of
Section 3.1 that such a union is closed. This means that the complement of this union is open. Note
that for any natural number q which is composite, q ∈ p∗ for some prime p∗ and, therefore, q is in
the union. Certainly all primes are in the union. Since all integers are either prime or composite
(other than 1 and -1), it follows that
2 ∪ 3 ∪ · · · ∪ p = X − {−1, 1}.
From this, we see that {−1, 1} is the complement of 2 ∪ 3 ∪ · · · ∪ p. But we said the complement
has to be open, and {−1, 1} is not open because it is not bi-infinite. This is a contradiction.
We conclude, then, that the number of prime numbers is not finite but is indeed infinite. ¤
5