3.5.2 Closed And Open Sets
I have been using the terminology
is a closed interval to mean it is an interval which
contains the two endpoints. However, there is a more general notion of what it means to
be closed. Similarly there is a general notion of what it means to be open. There is also a
concept of open. An example of an open set is
an interval which is missing its end
Definition 3.5.3 Let U be a set of points. A point p ∈ U is said to be an
interior point if whenever
is sufficiently small, it follows x ∈ U also. The set of
points, x which are closer to p than δ is denoted by
This symbol, B
is called an open ball of radius δ or an open interval. Thus a point,
p is an interior point of U if there exists δ >
0 such that p ∈ B
⊆ U. An open set
is one for which every point of the set is an interior point. Closed sets are those which
are complements of open sets. Thus H is closed means HC is open. Here HC
is more correctly denoted as ℝ ∖ H, and referrs to the set of all points in ℝ
not in H. In general, this is what is meant by the complement of a set. SC
denotes all points not in S which are in some given universal set containing
It consists of all points not in
. Thus it is (1
, all points which are
either to the right of 1 on the number line or to the left of 0 on the number line. Note
is an open set. Thus
is a closed set.
What is an example of an open set? The simplest example is an open ball.
Proposition 3.5.5 B
is an open set.
Proof: It is necessary to show every point is an interior point. Let x ∈ B
= δ −
. It follows
0 because it is given that
. Now consider
z ∈ B
and so z ∈ B
That is B
was arbitrary, this has
shown every point of the ball is an interior point. Thus the ball is an open set.
Definition 3.5.6 Let A be any nonempty set and let x be a point. Then
x is said to be a limit point of A if for every r > 0,B
contains a point of A
which is not equal to x.
Example 3.5.7 Consider A = ℕ, the positive integers. Then none of the points of
A is a limit point of A because if n ∈ A,B
contains no points of ℕ which
are not equal to n.
Example 3.5.8 Consider A =
, an open interval. If x ∈
⊆ A because if
< r, then showing y > a. A similar argument which you should provide shows y < b. Thus y ∈
and x is an interior point. Since x was arbitrary, this shows every point of
is an interior point and so
is open. Alternatively, you might notice
Theorem 3.5.9 Let A be a nonempty set. A point a is a limit point of A if
and only if there exists a sequence of distinct points of A,
which converges to
a, none of which equal a. Also a nonempty set A, is closed if and only if it contains
all its limit points.
Proof: Suppose first a is a limit point of A. There exists a1 ∈ B
. Now supposing distinct points, a1,
have been chosen such that none are
equal to a
and for each k ≤ n, ak ∈ B
Then there exists an+1 ∈ B
Because of the definition of rn+1,
is not equal to any of the other ak
for k < n
. Conversely, if there exists a sequence of distinct points of A
converging to a
none of which equal a,
contains infinitely many points of
since all are distinct. This establishes
the first part of the theorem.
Now consider the second claim. If A is closed then it is the complement of an open
set. Since AC is open, it follows that if a ∈ AC, then there exists δ > 0 such that
and so no point of AC
can be a limit point of A.
In other words, every
limit point of A
must be in A.
Conversely, suppose A contains all its limit points. Then AC does not contain any
limit points of A. It also contains no points of A. Therefore, if a ∈ AC, since it is not a
limit point of A, there exists δ > 0 such that B
contains no points of
itself is not in A
because a ∈ AC.
entirely contained in
Since a ∈ AC
was arbitrary, this shows every point
is an interior point and so AC
is open which shows that A
Corollary 3.5.10 Let A be a nonempty set and denote by A′ the set of limit
points of A. Then A ∪ A′ is a closed set and it is the smallest closed set containing
Proof: Is it the case that
is open? This is what needs to be shown if the
given set is closed. Let p
A ∪ A′.
Then since p
is neither in A
nor a limit point of A,
there exists B
. Therefore, B
also. This is
because if z ∈ B
and this smaller ball contains points of A since z is a limit point. This contradiction
shows that B
as claimed. Hence
is open because p
arbitrary point of
and A ∪ A′
is closed as claimed.
Now suppose C ⊇ A and C is closed. Then if p is a limit point of A, it follows from
Theorem 3.5.9 that there exists a sequence of distinct points of A converging to p. Since
C is closed, and these points of A are all in C, it follows that p ∈ C. Hence C ⊇ A ∪ A′.
Theorem 3.5.11 If K is sequentially compact and if H is a closed subset
of K then H is sequentially compact.
. Then since K
is sequentially compact, there is a subsequence,
which converges to a point,
x ∈ K.
then by Theorem 3.5.9
, which says
is open, it follows there exists B
such that this open ball contains no points of
However, this is a contradiction to having xnk → x
which requires xnk ∈ B
large enough. Thus x ∈ H
and this has shown H
is sequentially compact.
Thus every closed subset of a closed interval is sequentially compact. This is
equivalent to the following corollary in which a set is said to be bounded if it is contained
in some closed interval of finite length.
Corollary 3.5.12 Every closed and bounded set in ℝ is sequentially compact.
Proof: Let H be a closed and bounded set in ℝ. Then H is a closed subset of some
interval of the form
Therefore, it is sequentially compact. ■
In fact, one can go the other way.
Proposition 3.5.13 A nonempty set K ⊆ ℝ is sequentially compact if and
only if it is closed and bounded.
Proof: From the above corollary, if the set is closed and bounded, then it is
sequentially compact. Suppose now that K is sequentially compact. Why is it closed and
bounded? If it is not bounded, then you could pick
is sequentially compact, it follows that there is a subsequence,
The above indicates that for all j large enough, say j ≥ J,
1 and so by the
Therefore, the points
and so for all j,knj ∈
is bounded. Why must K
be closed? Suppose K
fails to contain p
is a limit point of K.
Then from Theorem 3.5.9
there exists a
sequence of distinct points of K,
such that lim
This is a
contradiction because the sequential compactness of K
requires the existence of a
such that lim
= q ∈ K.
because every subsequence of a convergent sequence converges to the limit of
the convergent sequence, and so p ∈ K
after all. This proves the proposition.