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.
Definition 4.7.3Let U be a set of points. A point p ∈ U is said to be an interiorpoint if whenever
|x− p|
is sufficiently small, it follows x ∈ U also. The set of points, x whichare closer to p than δ is denoted by
B (p,δ) ≡ {x ∈ F : |x − p| < δ} .
This symbol, B
(p,δ)
is called an open ball of radius δ. Thus a point, p is an interior point ofU if there exists δ > 0 such that p ∈ B
(p,δ)
⊆ U. An open set is one for which every pointof the set is an interior point.Closed sets are those which are complements of open sets.Thus H is closed means H^{C}is open.
What is an example of an open set? The simplest example is an open ball.
Proposition 4.7.4B
(p,δ)
is an open set.
Proof: It is necessary to show every point is an interior point. Let x ∈ B
∩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.
■
Corollary 4.7.9Let A be a nonempty set and denote by A^{′}the setof limit pointsof A. Then A ∪ A^{′}is a closed set and it is the smallest closed set containing A.
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}. Hence 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 4.7.8 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 4.7.10If K is sequentiallycompact and if H is a closed subset ofK then H is sequentially compact.
Proof: Let
{xn}
⊆ H. Then since K is sequentially compact, there is a subsequence,
{xnk}
which converges to a point, x ∈ K. If x
∕∈
H, then by Theorem 4.7.8, 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.
Corollary 4.7.11Every 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 4.7.12A nonempty set K ⊆ ℝ is sequentially compact if and only ifit 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→ ∞
However, this is impossible because this convergence can only take place if
{ }
knj
_{j=1}^{∞} is
bounded which it is not. Thus K is bounded. Why must it be closed? Suppose K fails to
contain p where p is a limit point of K. Then from Theorem 4.7.8 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
{pnk}
such that
lim_{k→∞}p_{nk} = q ∈ K. However, lim_{k→∞}p_{nk} = p and so p ∈ K after all. This proves the
proposition. ■
What about the sequentially compact sets in ℂ?
Definition 4.7.13A set S ⊆ ℂ is boundedif there is some r > 0 such thatS ⊆ B
(0,r)
.
Theorem 4.7.14Let H ⊆ ℂ.Then H is closed and bounded if and only if His sequentially compact.
Proof: Let H be a closed and bounded set in ℂ. Then H ⊆ B
(0,r)
for some r.
Therefore,
H ⊆ {x+ iy : x ∈ [− r,r] and y ∈ [− r,r]} ≡ Q
because if x + iy ∈ B
(0,r)
, then
∘x2-+-y2-
≤ r and so both
|x |
,
|y|
≤ r which is the same as
saying x ∈
[− r,r]
and y ∈
[− r,r]
. Now let
{x + iy }
n n
_{n=1}^{∞} be a sequence of points in H.
Then
{x }
n
is a sequence of points in
[− r,r]
and
{y }
n
is a sequence of points in
[− r,r]
.
It follows from Theorem 4.7.2 there exists a subsequence of
. It follows from the definition of distance in ℂ
that
xnkl + iynkl → x+ iy ∈ Q.
However, H is closed and so x + iy ∈ H.
Now suppose H is sequentially compact. Why is it closed and bounded? If it is not
bounded, then there exists
{zk}
⊆ H and this sequence is not bounded. Therefore, there is a
subsequence,
{znk}
such that lim_{k→∞}
|znk|
= ∞. However, since H is sequentially compact,
this requires that there exists a further subsequence which converges to something in H,
which is impossible because convergent sequences must be bounded. If H is not closed, then
there exists a point w which is a limit point of H but not in H. Since it is a limit point, there
exists a sequence
{zn}
⊆ H of distinct points such that lim_{n→∞}z_{n} = w. However, a
subsequence must converge to a point in H but this implies that w ∈ H because
every subsequence converges to w. Hence w ∈ H after all which is a contradiction.
■
What are some examples of closed and bounded sets in F?
Proposition 4.7.15Let D
(z,r)
denote the set of points,
{w ∈ F : |w − z| ≤ r}
Then D
(z,r)
is closed and bounded. Also any set of the form
[a,b]+i[c,d]
is closed and bounded. Thus sets D
(z,r)
and
[a,b]
+ i
[c,d]
are sequentially compact.
Proof: In case F = ℝ, there is nothing to show.
D (z,r) = [z − r,z + r]
Therefore, assume F = ℂ in what follows. Consider D
(z,r)
first. First note the set is bounded
because
D (z,r) ⊆ B (0,|z|+ 2r)
Here is why. Let x ∈ D
(z,r)
. Then
|x − z|
≤ r and so
|x | ≤ |x − z|+|z| ≤ r+ |z| < 2r+ |z|.
It remains to verify it is closed. Suppose then that y