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
(a,b)
, an interval which is missing its end
points.
Definition 3.5.3Let U be a set of points. A point p ∈ U is said to be aninterior point if whenever
|x − p|
is sufficiently small, it follows x ∈ U also. The set ofpoints, x which are closer to p than δ is denoted by
B (p,δ) ≡ {x ∈ ℝ : |x − p| < δ}.
This symbol, B
(p,δ)
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
(p,δ)
⊆ U. An open setis one for which every point of the set is an interior point.Closed sets are those whichare complements of open sets. Thus H is closed means H^{C}is open. Here H^{C}is more correctly denoted as ℝ ∖ H, and referrs to the set of all points in ℝnot in H. In general, thisis what is meant by the complement of a set. S^{C}denotes all points not in S which are in some given universal set containingS.
Example 3.5.4What is
[0,1]
^{C}?
It consists of all points not in
[0,1]
. Thus it is (1,∞) ∪
(− ∞,0)
, 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
that
[0,1]
^{C} is an open set. Thus
[0,1]
is a closed set.
What is an example of an open set? The simplest example is an open ball.
Proposition 3.5.5B
(p,δ)
is an open set.
Proof: It is necessary to show every point is an interior point. Let x ∈ B
. Since x was arbitrary, this has
shown every point of the ball is an interior point. Thus the ball is an open set.
■
Definition 3.5.6Let A be any nonempty set and let x be a point. Thenx is said to be a limit pointof A if for every r > 0,B
(x,r)
contains a point of Awhich is not equal to x.
Example 3.5.7Consider A = ℕ, the positive integers. Then none of the points ofA is a limit point of A because if n ∈ A,B
(n,1∕10)
contains no points of ℕ whichare not equal to n.
Example 3.5.8Consider A =
(a,b)
, an open interval. If x ∈
(a,b)
, let
r = min (|x− a|,|x− b|).
Then B
(x,r)
⊆ A because if
|y − x|
< r, then
y − a = y− x + x− a ≥ x− a − |y− x|
= |x− a|− |y − x| > |x − a|− r ≥ 0
showing y > a. A similar argument which you should provide shows y < b. Thus y ∈
(a,b)
and x is an interior point. Since x was arbitrary, this shows every point of
(a,b)
is an interior point and so
(a,b)
is open. Alternatively, you might noticethat
(a + b b− a)
(a,b) = B -----,----- .
2 2
Theorem 3.5.9Let A be a nonempty set.A point a is a limit point of A ifand only if there exists a sequence of distinct points of A,
{an}
which converges toa, none of which equal a. Also a nonempty set A, is closed if and only if it containsall its limit points.
Proof: Suppose first a is a limit point of A. There exists a_{1}∈ B
(a,1)
∩A such that
a_{1}≠a. Now supposing distinct points, a_{1},
⋅⋅⋅
,a_{n} have been chosen such that none are
equal to a and for each k ≤ n, a_{k}∈ B
(a,1∕k)
, let
{ 1 }
0 < rn+1 < min -----,|a − a1|,⋅⋅⋅,|a− an| .
n + 1
Then there exists a_{n+1}∈ B
(a,rn+1)
∩A with a_{n+1}≠a. Because of the definition of r_{n+1},a_{n+1} is not equal to any of the other a_{k} for k < n + 1. Also since
|a− am |
< 1∕m, it
follows lim_{m→∞}a_{m} = a. Conversely, if there exists a sequence of distinct points of A
converging to a none of which equal a, then B
(a,r)
contains all a_{n} for n large enough.
Thus B
(a,r)
contains infinitely many points of A 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 A^{C} is open, it follows that if a ∈ A^{C}, then there exists δ > 0 such that
B
(a,δ)
⊆ A^{C} and so no point of A^{C} 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 A^{C} does not contain any
limit points of A. It also contains no points of A. Therefore, if a ∈ A^{C}, since it is not a
limit point of A, there exists δ > 0 such that B
(a,δ)
contains no points of A different
than a. However, a itself is not in A because a ∈ A^{C}. Therefore, B
(a,δ)
is
entirely contained in A^{C}. Since a ∈ A^{C} was arbitrary, this shows every point
of A^{C} is an interior point and so A^{C} is open which shows that A is closed.
■
Corollary 3.5.10Let A be a nonempty set and denote by A^{′}the setof limitpoints of A. Then A ∪ A^{′}is a closed set and it is the smallest closed set containingA.
Proof: Is it the case that
(A ∪ A′)
^{C} 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
(p,r)
such that B
(p,r)
∩ A = ∅. Therefore, B
(p,r)
∩ A^{′} = ∅ also. This is
because if z ∈ B
(p,r)
∩ A^{′}, then
B (z,r − |p− z|) ⊆ B (p,r)
and this smaller ball contains points of A since z is a limit point. This contradiction
shows that B
(p,r)
∩ A^{′} = ∅ as claimed. Hence
′
(A∪ A )
^{C} is open because p was an
arbitrary point of
′
(A∪ A )
^{C}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 3.5.11If K is sequentiallycompact and if H is a closed subsetof K then H is sequentially compact.
Proof: Let
{xn}
⊆ H. Then since K is sequentially compact, there is a subsequence,
{xn}
k
which converges to a point, x ∈ K. If x
∕∈
H, then by Theorem 3.5.9, which says
H^{C} is open, it follows there exists B
(x,r)
such that this open ball contains no points of
H. However, this is a contradiction to having x_{nk}→ x which requires x_{nk}∈ B
(x,r)
for
all k 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 3.5.12Every closed and bounded set in ℝ is sequentiallycompact.
Proof: Let H be a closed and bounded set in ℝ. Then H is a closed subset of some
interval of the form
[a,b]
. Therefore, it is sequentially compact. ■
In fact, one can go the other way.
Proposition 3.5.13A nonempty set K ⊆ ℝ is sequentially compact if andonly 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
{kn}
_{n=1}^{∞} such that
|kn|
≥ n. Since
K is sequentially compact, it follows that there is a subsequence,
{kn }
j
which
satisfies
lim knj = k ∈ K
j→∞
The above indicates that for all j large enough, say j ≥ J,
| |
|k − knj|
< 1 and so by the
triangle inequality
| |
|knj| ≤ |k|+ 1
Therefore, the points
||knj||
< r where
r = max {|k|+ 1,||k ||+ 1 for j ≤ J }
nj
and so for all j,k_{nj}∈
[− r,r]
. Thus K is bounded. Why must K be closed? Suppose K
fails to contain p where p is a limit point of K. Then from Theorem 3.5.9 there exists a
sequence of distinct points of K,
{pn}
such that lim_{n→∞}p_{n} = p
∕∈
K. This is a
contradiction because the sequential compactness of K requires the existence of a
subsequence
{pn }
k
such that lim_{k→∞}p_{nk} = q ∈ K. However, lim_{k→∞}p_{nk} = p
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.
■