We now investigate some properties that subsets of RR and CC
may possess. We will define “closed sets,” “open sets,” and “limit points” of sets.
These notions are the rudimentary notions of what is called topology.
As in earlier definitions, these topological ones will be enlightening when we come to continuity.
- Definition 1:
Let SS be a subset of C.C.
A complex number xx
is called a limit point of SS if there exists a sequence {xn}{xn} of elements of SS
such
that x=limxn.x=limxn.
A set S⊆CS⊆C is called closed
if every limit point of SS belongs to S.S.
Every limit point of a set of real numbers is a real number.
Closed intervals [a,b][a,b] are examples of closed sets in R,R, while open intervals and half-open intervals may not be closed sets.
Similarly, closed disks Br(c)¯Br(c)¯ of radius rr around a point cc in CC,
and closed neighborhoods N¯r(S)N¯r(S) of radius rr around a set S⊆C,S⊆C,
are closed sets, while the open disks or open neighborhoods are not closed sets.
As a first example of a limit point of a set, we give
the following exercise.
Let SS be a nonempty bounded set of real numbers, and let M=supS.M=supS.
Prove that there exists a sequence {an}{an} of elements of SS such that M=liman.M=liman.
That is, prove that the supremum of a bounded set of real numbers is a limit point of that set.
State and prove an analogous result for infs.
HINT: Use (Reference), and let ϵϵ run through the numbers 1/n.1/n.
- Suppose SS is a set of real numbers, and that z=a+bi∈Cz=a+bi∈C
with b≠0.b≠0. Show that zz is not a limit point of S.S.
That is, every limit point of a set of real numbers is a real number.
HINT: Suppose false; write a+bi=limxn,a+bi=limxn, and make
use of the positive number |b|.|b|.
- Let cc be a complex number, and let
S=B¯r(c)S=B¯r(c) be the set of all z∈Cz∈C for which |z-c|≤r.|z-c|≤r.
Show that SS is a closed subset of CC.
HINT: Use part (b) of (Reference).
- Show that the open disk Br(0)Br(0) is not a closed set in CC by finding
a limit point of Br(0)Br(0) that is not in Br(0).Br(0).
- State and prove results analogous to parts b and c for
intervals in R.R.
- Show that every element xx of a set SS is a limit point of S.S.
- Let SS be a subset of CC, and let xx be a complex number.
Show that xx is not a limit point of SS if and only if
there exists a positive number ϵϵ such that if |y-x|<ϵ,|y-x|<ϵ, then yy is not in S.S.
That is, S∩Bϵ(x)=∅.S∩Bϵ(x)=∅.
HINT:
To prove the “ only if” part, argue by contradiction,
and use the sequence {1/n}{1/n} as ϵϵ's.
- Let {an}{an} be a sequence of complex numbers, and let SS
be the set of all the anan's.
What is the difference between a cluster point of the sequence {an}{an} and a limit point of the set S?S?
- (h) Prove that the cluster set of a sequence is a closed set.
HINT: Use parts (e) and (f).
- Show that the set QQ of all rational numbers is not a closed set.
Show also that the set of all irrational numbers is not a closed set.
- Show that if SS is a closed subset of RR that contains Q,Q, then SS must equal all of R.R.
Here is another version of the Bolzano-Weierstrass Theorem, this time
stated in terms of closed sets rather than bounded sequences.
Let SS be a bounded and closed subset of CC.
Then every sequence {xn}{xn} of elements of SS has a subsequence that converges to an element of S.S.
Let {xn}{xn} be a sequence in S.S.
Since SS is bounded, we know by (Reference) that there exists a subsequence {xnk}{xnk} of {xn}{xn} that converges to
some
number x.x.
Since each xnkxnk belongs to S,S, it follows that xx is a limit point of S.S.
Finally, because SS is a closed subset of C,C, it then follows that x∈S.x∈S.
We have defined the concept of a closed set.
Now let's give the definition of an open set.
- Definition 2:
Let SS be a subset of C.C. A point x∈Sx∈S is
called an interior point of SS if there exists an ϵ>0ϵ>0
such that the open disk Bϵ(x)Bϵ(x) of radius ϵϵ around xx is entirely contained in S.S.
The set of all interior points of SS is denoted by S0S0
and we call S0S0 the interior of S.S.
A subset SS of CC
is called an open subset of CC
if every point of SS is an interior point of S;S;
i.e., if S=S0.S=S0.
Analogously, let SS be a subset of R.R. A point x∈Sx∈S is
called an interior point of SS if there exists an ϵ>0ϵ>0
such that the open interval (x-ϵ,x+ϵ)(x-ϵ,x+ϵ)
is entirely contained in S.S.
Again, we denote the set of all interior points of SS by S0S0
and call S0S0 the interior of S.S.
A subset SS of RR
is called an open subset of RR
if every point of SS is an interior point of S;S;
i.e., if S=S0.S=S0.
- Prove that an open interval (a,b)(a,b) in RR is an
open subset of RR; i.e., show that every point of (a,b)(a,b) is an interior point of (a,b).(a,b).
- Prove that any disk Br(c)Br(c) is an open subset of C.C.
Show also that the punctured diskBr'(c)Br'(c) is an open set,
where Br'(c)={z:0<|z-c|<r},Br'(c)={z:0<|z-c|<r}, i.e., evrything in the disk Br(c)Br(c) except the central point c.c.
- Prove that the neighborhood Nr(S)Nr(S) of radius rr around a set SS
is an open subset of C.C.
- Prove that no nonempty subset of RR is an open subset of C.C.
- (e) Prove that the set QQ of all rational numbers is not an open subset of R.R.
We have seen in part (a) of Exercise 3 that QQ is
not a closed set. Consequently it is an example of a set that is neither open nor closed.
Show that the set of all irrational numbers is neither open nor closed.
We give next a useful application of the Bolzano-Weierstrass Theorem, or more precisely an
application of Theorem 1.
This also provides some insight into the structure of open sets.
Let SS be a closed and bounded subset of C,C, and suppose
SS is a subset of an open set U.U.
Then there exists an r>0r>0 such that the neighborhood Nr(S)Nr(S) is contained in U.U.
That is, every open set containing a closed and bounded set SS actually contains a neighborhood of S.S.
If SS is just a singleton {x},{x}, then this theorem is asserting nothing more than
the fact that xx is in the interior of U,U,
which it is if UU is an open set.
However, when SS is an infinite set, then the result is more subtle.
We argue by contradiction. Thus, suppose there is no such r>0r>0
for which Nr(S)⊆U.Nr(S)⊆U.
then for each positive integer nn there must be a point xnxn that is not in U,U, and a corresponding point yn∈S,yn∈S,
such that |xn-yn|<1/n.|xn-yn|<1/n. Otherwise, the number r=1/nr=1/n would satisfy the claim of the theorem.
Now, because the ynyn's all belong to S,S, we know from Theorem 1 that a subsequence {ynk}{ynk} of the sequence {yn}{yn}
must converge to a number y∈S.y∈S. Next, we see that
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(1)and this quantity tends to 0. Hence, the subsequence {xnk}{xnk} of the sequence {xn}{xn} also converges to y.y.
Finally, because yy belongs to SS and hence to the open set U,U, we know that there must exist an ϵ>0ϵ>0 such that the entire
disk Bϵ(y)⊆U.Bϵ(y)⊆U.
Then, since the subsequence {xnk}{xnk} converges
to y,y, there must exist anknk such that |xnk-y|<ϵ,|xnk-y|<ϵ, implying that
xnk∈Bϵ(y),xnk∈Bϵ(y), and hence belongs to
U.U. But this is our contradiction, because all of the xnxn's were not in U.U.
So, the theorem is proved.
We give next a result that clarifies to some extent the connection between open sets and closed sets.
Always remember that there are sets that are neither open nor closed, and just because a set is not open
does not mean that it is closed.
A subset SS of CC (RR) is open
if and only if its complement S˜=C∖SS˜=C∖S
(R∖SR∖S) is closed.
First, assume that SS is open, and let us show that S˜S˜ is closed.
Suppose not. We will derive a contradiction.
Suppose then that there is a sequence {xn}{xn} of elements of S˜S˜ that converges to a number
xx that is not in S˜;S˜; i.e., xx is an element of S.S.
Since every element of SS is an interior point of S,S, there
must exist an ϵ>0ϵ>0 such that the entire disk Bϵ(x)Bϵ(x) (or interval (x-ϵ,x+ϵ)(x-ϵ,x+ϵ)) is a subset of S.S.
Now, since x=limxn,x=limxn, there must exist anNN such that |xn-x|<ϵ|xn-x|<ϵ
for every n≥N.n≥N. In
particular, |xN-x|<ϵ;|xN-x|<ϵ; i.e., xNxN belongs to Bϵ(x)Bϵ(x) (or (x-ϵ,x+ϵ)(x-ϵ,x+ϵ)).
This implies that xN∈S.xN∈S. But xN∈S˜,xN∈S˜, and this is a contradiction.
Hence, if SS is open, then S˜S˜ is closed.
Conversely, assume that S˜S˜ is closed, and let us show that SS must be open.
Again we argue by contradiction.
Thus, assuming that SS is not open, there must exist a point x∈Sx∈S
that is not an interior point of S.S. Hence, for every ϵ>0ϵ>0
the disk Bϵ(x)Bϵ(x) (or interval (x-ϵ,x+ϵ)(x-ϵ,x+ϵ)) is not entirely contained in S.S.
So, for each positive integer n,n, there must exist a point
xnxn such that |xn-x|<1/n|xn-x|<1/n and xn∉S.xn∉S.
It follows then that x=limxn,x=limxn, and that each xn∈S˜.xn∈S˜.
Since S˜S˜ is a closed set, we must have that x∈S˜.x∈S˜.
But x∈S,x∈S, and we have arrived at the desired contradiction.
Hence, if S˜S˜ is closed, then SS is open,
and the theorem is proved.
The theorem below, the famous Heine-Borel Theorem,
gives an equivalent and different description of closed and bounded sets.
This description is in terms of open sets, whereas the original definitions
were interms of limit points.
Any time we can find two very different descriptions of the same phenomenon, we have
found something useful.
- Definition 3:
Let SS be a subset of CC (respectively R).R). By an open cover
of SS we mean a sequence {Un}{Un} of open subsets of CC (respectively RR)
such that S⊆∪Un;S⊆∪Un;
i.e., for every x∈Sx∈Sthere exists an nn such that x∈Un.x∈Un.
A subset SS of CC (respectively RR)
is called compact,
or is said to satisfy the
Heine-Borel property, if every open cover of SS
has a finite subcover. That is, if {Un}{Un}
is an open cover of S,S, then there exists an integer NN such that
S⊆∪n=1NUn.S⊆∪n=1NUn.
In other words, only a finite number of the open sets are necessary to cover S.S.
REMARK
The definition we have given here for a set being compact is
a little less general from the one found in books on topology.
We have restricted the notion of an open cover to be a sequence of
open sets, while in the general setting an open
cover is just a collection of open sets.
The distinction between a sequence of open sets and a collection of open sets
is genuine in general topology, but it can be
disregarded in the case of the topological spaces RR and C.C.
A subset SS of CC (respectively RR) is compact
if and only if it is a closed and bounded set.
We prove this theorem for subsets SS of C,C,
and leave the proof for subsets of RR to the exercises.
Suppose first that S⊆CS⊆C is compact, i.e., satisfies the Heine-Borel property.
For each positive integer n,n, define UnUn to be the
open set Bn(0).Bn(0). Then S⊆∪Un,S⊆∪Un, because C=∪Un.C=∪Un.
Hence, by the Heine-Borel property, there must exist an NN such
that S⊆∪n=1NUn.S⊆∪n=1NUn. But then S⊆BN(0),S⊆BN(0),
implying that SS is bounded.
Indeed, |x|≤N|x|≤N for all x∈S.x∈S.
Next, still assuming that SS is compact, we will show that SS is closed by showing that S˜S˜ is open.
Thus, let xx be an element of S˜.S˜. For each positive integer n,n, define
UnUn to be the complement of the closed set B1/n(x)¯.B1/n(x)¯.
Then each UnUn is an open set by Theorem 2.12, and we claim that {Un}{Un}
is an open cover of S.S. Indeed, if y∈S,y∈S, then y≠x,y≠x, and
|y-x|>0.|y-x|>0. Choose an nn so that 1/n<|y-x|.1/n<|y-x|.
Then y∉B1/n(x)¯,y∉B1/n(x)¯, implying that y∈Un.y∈Un.
This proves our claim that {Un}{Un} is an open cover of S.S.
Now, by the Heine-Borel property,
there exists an NN such that S⊆∪n=1NUn.S⊆∪n=1NUn.
But this implies that for every z∈Sz∈S we must have
|z-x|≥1/N,|z-x|≥1/N, and this implies that the disk
B1/N(x)B1/N(x) is entirely contained in S˜.S˜. Therefore, every element xx of S˜S˜
is an interior point of S˜.S˜.
So, S˜S˜ is open, whence SS is closed.
This finishes the proof that compact sets are necessarily closed and bounded.
Conversely, assume that SS is both closed and bounded.
We must show that SS satisfies the Heine-Borel property.
Suppose not. Then, there exists an open cover
{Un}{Un} that has no finite subcover. So, for each positive integer nn there must exist an element xn∈Sxn∈S
for which xn∉∪k=1nUk.xn∉∪k=1nUk. Otherwise, there would be a finite subcover.
By Theorem 1, there exists a subsequence {xnj}{xnj} of {xn}{xn} that converges
to an element xx of S.S. Now, because {Un}{Un} is an open cover of S,S,
there must exist an NN such that x∈UN.x∈UN.
Because UNUN is open, there exists an ϵ>0ϵ>0 so that the
entire disk Bϵ(x)Bϵ(x) is contained in UN.UN.
Since x=limxnj,x=limxnj, there exists a JJ so that |xnj-x|<ϵ|xnj-x|<ϵ
if j≥J.j≥J. Therefore, if j≥J,j≥J, then xnj∈UN.xnj∈UN.
But the sequence {nj}{nj} is strictly increasing, so that there exists a j'≥Jj'≥J such that nj'>N,nj'>N, and by the choice of the
point xnj',xnj', we know that xnj'∉∪k=1NUk.xnj'∉∪k=1NUk.
We have arrived at a contradiction, and so the second half of the theorem is proved.
- Prove that the union A∪BA∪B of two open sets is open and the intersection A∩BA∩B is also open.
- Prove that the union A∪BA∪B of two closed sets is closed and the intersection A∩BA∩B is also closed.
HINT: Use Theorem 3 and the set equations
A∪B˜=A˜∩B˜,A∪B˜=A˜∩B˜, and
A∩B˜=A˜∪B˜.A∩B˜=A˜∪B˜.
These set equations are known as Demorgan's Laws.
- Prove that the union A∪BA∪B of two bounded sets is bounded and the intersection A∩BA∩B is also bounded.
- Prove that the union A∪BA∪B of two compact sets is compact and the intersection A∩BA∩B is also compact.
- Prove that the intersection of a compact set and a closed set is compact.
- Suppose SS is a compact set in CC and rr is a positive real number.
Prove that the closed neighborhood N¯r(S)N¯r(S) of radius rr around SS is compact.
HINT: To see that this set is closed, show that its coplement is open.
