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Transcript
More Open Sets and Topological Compactness
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
September 13, 2016
Outline
More Set Theory
Topological Compactness
There are more things we can talk about with sets. So let’s get started.
You’ll see that the way we prove propositions in set theory is very
different from the way we do this in other parts of this class. All part of
your growth!
Example
Let A and B be open sets. Then A ∪ B is open too.
Solution
Remember A ∪ B = {c : a ∈ A and / or b ∈ B}. To show this result, we
must show an arbitrary c in A ∪ B is an interior point. Now for such a c
it is clear c is in A only or it is in B only or it is in both. All that matters
here is that c is in one or the other. If c ∈ A, then c is an interior point
and there is a Br (c) ⊂ A ⊆ A ∪ B. And if c ∈ B, again c is an interior
point and there is a Br (c) ⊂ B ⊆ A ∪ B. So in all cases, c is an interior
point and so A ∪ B is open.
A countable union is one where the index set is an RII of Z. Usually N is
what we use, but it could be such a subset.
Example
The countable union of open sets is open.
Solution
Let our RII set be N just for convenience of notation and let the
individual open sets here be Oi . We let A = ∪∞
n=1 On = ∪n∈N On denote
the countable union. Let x be in the union. Then x is in at least one OQ .
Since OQ is open, x is an interior point of OQ . So there is a
Br (x) ⊂ OQ . But OQ ⊆ ∪n∈N On . Hence, x is an interior point of the
countable union. Since x was arbitrary, the countable union is open.
Comment
For a countable union of sets, we usually just write ∪n On .
Example
Unions need not be over countable index sets. A good example is this.
Let S = [1, 3) which has an infinite number of points. At each x ∈ S,
pick the radius rx > 0 and consider the set A = ∪x∈S Brx . This index set
is not an RII so indeed is an example of what is called a uncountable set.
To be precise, a countable set Ω is one for which there is a map f which
assigns each element in Ω one and only one element of N. A RII is
countable because nk is assigned to k in N and this assignment is unique.
Hence there is a 1 − 1 and onto map from the RII to N.
Example
The rational numbers are countable too. We will do the construction for
positive fractions only but you should be able to see how to extend it to
negative fractions also. We can set up the map like this:
Example
(Continued):
Write down all the fraction in a table:
1/1
2/1
3/1
4/1
..
.
1/2
2/2
3/2
4/2
..
.
1/3
2/3
3/3
4/3
..
.
1/4
2/4
3/4
4/4
..
.
1/5
2/5
3/5
4/5
..
.
1/6
2/6
3/6
4/6
..
.
1/6
2/6
3/6
4/6
..
.
...
...
...
...
..
.
1/n
2/n
3/n
4/n
..
.
...
...
...
...
..
.
n/1
..
.
n/2 n/3
..
..
.
.
n/4
..
.
n/5
..
.
n/6
..
.
n/6
..
.
...
..
.
n/n
..
.
...
..
.
Example
(Continued):
The mapping is then defined by the numbers in parenthesis chosen in
diagonal sweeps:
1/1 (1)
2/1 (3)
3/1 (4)
4/1 (10)
..
.
n/1
..
.
1/2 (2) 1/3 (6)
2/2 (5) 2/3 (8)
3/2 (9)
3/3
4/2
4/3
..
..
.
.
n/2
n/3
..
..
.
.
1/4 (7)
2/4
3/4
4/4
..
.
1/5
2/5
3/5
4/5
..
.
1/6
2/6
3/6
4/6
..
.
1/6
2/6
3/6
4/6
..
.
...
...
...
...
..
.
1/n
2/n
3/n
4/n
..
.
...
...
...
...
..
.
n/4
..
.
n/5
..
.
n/6
..
.
n/6
..
.
...
..
.
n/n
..
.
...
..
.
Example
(Continued):
eliminating the fractions, it is easier to see the pattern:
(1)
(2)
(6)
(7)
(3)
(5)
(8)
(4)
(9)
(10)
(12)
..
.
(13)
..
.
(14)
..
.
(11)
(15) . . .
..
.
...
...
...
...
I
So the rationals Q are countable too. We often call this countably
infinite to distinguish it from the case where the index set is
countably finite like {1, 2, 3, 4, 5}.
I
So [1, 5] would not be countable as it contains lots of numbers
which are not fractions.
Example
The uncountable union of open sets is open.
Solution
Let the uncountable index set be Ω. and let the individual open sets here
be Oα where α is a member of the index set Ω. We let
A = ∪α∈Ω Oα = ∪α Oα . denote the uncountable union. Let x be in the
union. Then x is in at least one Oβ . Since Oβ is open, x is an interior
point of Oβ . So there is a Br (x) ⊂ Oβ . But Oβ ⊆ ∪α Oα . Hence, x is
an interior point of the uncountable union. Since x was arbitrary, the
uncountable union is open.
Example
The intersection of a finite number of open sets is open too.
Solution
Let A = ∩N
n=1 On be a finite intersection of open sets. Let x be in this
intersection. Then x is an interior point of all N sets. Hence, we have N
circles: Br1 (x) ⊂ O1 through BrN (x) ⊂ ON . If we let r = min{r1 , . . . , rN },
then Br (x) is contained in each of the N sets. Hence A is open.
Example
The intersection of a countable number of open sets need not be open.
Solution
Take the sets (1 − 1/n, 4 + 1/n). We claim ∩n (1 − 1/n, 4 + 1/n) = [1, 4],
a closed set. To prove these sets are equal, we first show
∩n (1 − 1/n, 4 + 1/n) ⊂ [1, 4] and then [1, 4] ⊆ ∩n (1 − 1/n, 4 + 1/n).
(∩n (1 − 1/n, 4 + 1/n) ⊂ [1, 4]):
Let x ∈ ∩n (1 − 1/n, 4 + 1/n). Then, x ∈ (1 − 1/n, 4 + 1/n) ⊂ [1, 4] for
all n. Thus, x ∈ [1, 4].
([1, 4] ⊆ ∩n (1 − 1/n, 4 + 1/n)):
Let x ∈ [1, 4]. Then x ∈ (1 − 1/n, 4 + 1/n) for all n implying
x ∈ ∩n (1 − 1/n, 4 + 1/n).
Comment
So the property of being open can be violated if a countably infinite
number of operations are used.
We need lots of small results in this kind of thinking. Here is one:
Example
∩N
n=1 Bn
C
C
= ∪N
n=1 Bn for any sets.
Solution C
∩N
⊆ ∪N
BCn
n=1 Bn
n=1
C
If x ∈ ∩N
, then x 6∈ ∩N
n=1 Bn
n=1 Bn . That tells us x 6∈ Bn for at least
C
one k. Hence, x ∈ BCk implying x ∈ ∪N
n=1 Bn .
C
C
N
∪N
B
⊆
∩
B
.
n
n=1 n
n=1
C
C
If x ∈ ∪N
n=1 Bn , there is an index Q with x ∈ BQ . Thus x 6∈ BQ which
C
N
N
tells us x 6∈ ∩n=1
Bn . Hence, x ∈ ∩n=1
Bn .
Example
The intersection of a finite number of closed sets is closed too.
Solution
Let A = ∩N
n=1 Dn be a finite intersection of closed sets. We now know
C
C
C
= ∪N
AC = ∩N
n=1 Dn
n=1 Dn . Since each Dn is closed, Dn is open. We
also know the finite union of open sets is open. Then AC is open
implying A is closed.
Comment
C
∞
C
We can also prove (∩∞
n=1 Bn ) = ∪n=1 Bn for any sets Bn and prove the
intersection of a countable number of closed sets is still closed.
Example
The union of a countable number of closed sets need not be closed.
Solution
We claim ∪n [1 + 1/n, 4 − 1/n] = (1, 4), an open set.
(∪n (1 + 1/n, 4 − 1/n) ⊂ (1, 4) ):
Let x ∈ ∪n [1 + 1/n, 4 − 1/n]. Then, x ∈ [1 + 1/Q, 4 − 1/Q] for some Q.
Thus, the distance from x to 1 or 4 is positive. So x ∈ (1, 4).
( (1, 4) ⊆ ∪n [1 + 1/n, 4 − 1/n]):
If x ∈ (1, 4), x is an interior point and so there is Br (x) ⊂ (1, 4). That is,
1 < x − r < x < x + r < 4. Let the distance from x − r to 1 be d1 and
the distance from x + r to 4 be d2 . Let d = (1/2) min{d1 , d2 }. Choose
N so that 1/N < d. Then x ∈ [1 + 1/N, 4 − 1/N] and so
x ∈ ∪n [1 + 1/n, 4 − 1/n].
Comment
The property of closed can be violated using countably infinite operations.
Definition
We say a collection of open sets U = {Oα : α ∈ Ω} indexed by an
arbitrary set Ω ( so it can be finite, countably infinite or uncountable)
is a open cover of a set A if A ⊆ ∪α∈Ω Oα .
We say the collection {Oα1 , . . . , OαN } from U , for some natural
number N, is a finite subcover or fsc of A if A ⊆ ∪N
n=1 Oαn .
We are now ready for the big new idea!
Definition
We say a set A is topologically compact if every open cover of A has
a finite subcover.
We need to show this idea in our setting < is equivalent to sequential
compactness.
Theorem
A is topologically compact ⇒ A is closed and bounded.
Proof
(⇒):
Let’s assume A is topologically compact. If we take the union of all
Br (x) for a fixed radius r for all x in S. this is an open cover. Since A
is topologically compact, this cover has a fsc which contains A. We
can express the fsc as V = {Br (x1 ), . . . , Br (xN )} for some N. Now let
x ∈ A. Then x ∈ Br (xj ) for some 1 ≤ j ≤ N. Thus,
|x − xj | < r ⇒ |x| < r + |xj | using the backwards triangle inequality.
And in turn, r + |xj | ≤ max{r + |x1 |, . . . , r + |xN |} = B. So |x| ≤ B
for all x ∈ A implying A is bounded.
To show A is closed, we will show AC is open. Let x ∈ AC . Given any
y ∈ A, since x 6= y , the two points are separated by a distance d xy .
Thus, Bd xy /2 (y ) and Bd xy /2 (x) have no points in common.
Proof
The collection of all the circles Bd xy /2 (y ) for y ∈ A gives an open cover
U of A which must have a fsc, V = {Bd1xy (y1 ), . . . , BdNxy (yN )}, so that
xy
A ⊆ ∪N
n=1 Bdn (yn ).
xy
xy
Let W = ∩N
n=1 Bdn (x). The circles Bdn (x) are all separate ( we say
disjoint ) from Bdnxy (y ). Hence, points in the intersection are outside the
union of the fsc V . Then W is open circle as it is a finite intersection of
open circles and it is outside of the union over the fsc V which contains
A. Since each of these sets is contained in AC , we see W ⊂ AC implying
x is an interior point of AC . Since x was arbitrary, we see AC is open.
Hence A is closed.
Theorem
The closed and bounded finite interval [a, b], a < b, is topologically
compact.
Proof
This proof is similar to how we proved the Bolzano Weierstrass
Theorem (BWT). So look at the similarities!
Assume this is false and let J0 = [a, b]. Label this interval as
J0 = [α0 , β0 ]; i.e. α0 = a and β0 = b. The length of J0 is b − a = B
which we will call `0 . Assume there is an open cover U which does
not have a fsc.
Now divide J1 into two equal pieces
b−a
{x0 = α0 , x1 = α0 + 1 b−a
2 , x2 = a + 2 2 = b}. We have
[α0 , β0 ] = [x0 , x1 ] ∪ [x1 , x2 ] At least one of these intervals can not be
covered by a fsc as otherwise the entire interval [α0 , β0 ] has a fsc
which we have assumed is not true. Call this interval J1 and note the
length of J1 is `1 = (1/2)`0 = B/2.
Proof
Label this interval as J1 = [α1 , β1 ].
We can continue in this way by induction. Assume we have constructed
the interval Jn = [αn , βn ] in this fashion with length
`n = (1/2)`n−1 = B/2n and there is no fsc of Jn . Now divide Jn into 2
equal pieces as usual {x0 = αn , x1 = αn + 1 `n , x2 = αn + 2 `n = βn }.
These points determine 2 subintervals and at least one of them can not
be covered by a fsc. Call this interval Jn+1 = [αn+1 , βn+1 ] which has
length `n+1 = (1/2)`n like required.
We can see each Jn+1 ⊂ Jn by the way we have chosen the intervals and
their lengths satisfy `n → 0. Using the same sort of arguments that we
used in the proof of the BWT for Sequences, we see there is a unique
point z which is the sup(αn ) = inf(βn ). Another way to say this in our
new set language is that ∩∞
n=1 [αn , βn ] = {z}!
Proof
Since z in in J1 , there is a O1 in U with z ∈ O1 because U covers A.
Also, since O1 is open, there is a circle Br1 (z) ⊂ O1 . Now choose K so
that `K < `1 . Then, Jk is contained in O1 for all k > K . This says O1 is
a fsc of Jk which contradicts our construction process. Hence, our
assumption that there is an open cover with no fsc is wrong and we have
[a, b] is topologically compact.
Homework 12
12.1 If A and B are nonempty sets, show all the details of the proof for
(A ∪ B)C = AC ∩ B C .
12.2 Find an open cover with no fsc for the set (1, 3). Explain why this
shows (1, 3) is not topologically compact.
12.3 If A and B are nonempty sets, show all the details of the proof for
(A ∩ B)C = AC ∪ B C .