Definition B.1.1Let U be a set of points in F^{p}. A point, p ∈ U is said to be aninterior point of U 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 ∈ V : ||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 point ofthe set is an interior point.Closed sets are those which are complements of open sets. ThusH is closed means H^{C}is open.
The first question to consider is whether an “open ball” is actually an open
set.
Proposition B.1.2An open ball in F^{p}is an open set.
Proof: Let z ∈ B
(x,r)
. Is there a ball B
(z,δ)
⊆ B
(x,r)
? Yes there is. Consider
B
(z,r− ||x − z||)
. If y is in this last ball, then
||y − x|| ≤ ||y − z||+ ||z − x|| < r − ||x− z||+||z− x|| = r.
Thus every point of the ball is indeed an interior point and this shows the open ball is indeed
open. ■
Note that the definition of open and closed does not depend on the choice of the norm
because of ∗ which implies that in ℝ^{n}, B
(x,r)
⊆ B_{∞}
(x,r)
⊆ B
( √ - )
x, pr
. It turns out that all
norms are equivalent. Thus when we mention open set or closed set, it will end up making no
difference which norm is considered. In particular it does not matter whether the norm is
|⋅|
or
∥⋅∥
. However, notice that
n∏
B ∞ (p,r) = (pi − r,pi + r)
i=1
a product of open intervals. This is especially convenient.
Theorem B.1.3The intersection of any finite collection of open sets is open.The union of any collection of open sets is open. The intersection of any collection ofclosed sets is closed and the union of any finite collection of closed sets is closed.
Proof:To see that any union of open sets is open, note that every point p of the union is
in at least one of the open sets U. Therefore, it is an interior point of U and hence an interior
point of the entire union.
Now let
{U1,⋅⋅⋅,Um }
be some open sets and suppose p ∈∩_{k=1}^{m}U_{k}. Then there
exists r_{k}> 0 such that B
(p,rk)
⊆ U_{k}. Let 0 < r ≤ min
(r1,r2,⋅⋅⋅,rm)
. Then
B
(p,r)
⊆∩_{k=1}^{m}U_{k} and so the finite intersection is open. Note that if the finite intersection
is empty, there is nothing to prove because it is certainly true in this case that
every point in the intersection is an interior point because there aren’t any such
points.
Suppose
{H1,⋅⋅⋅,Hm }
is a finite set of closed sets. Then ∪_{k=1}^{m}H_{k} is closed if its
complement is open. However,
(∪mk=1Hk)C = ∩mk=1HCk ,
a finite intersection of open sets which is open by what was just shown.
Next let C be a set consisting of closed sets. Then
{ }
(∩ C)C = ∪ HC : H ∈ C ,
a union of open sets which is therefore open by the first part of the proof. Thus ∩C is closed.
This proves the theorem. ■
Next there is the concept of a limit point which gives another way of characterizing closed
sets.
Definition B.1.4Let A be any nonempty set and let p be a point. Then p issaid to be a limit pointof A if for every r > 0,B
(p,r)
contains a point of A which isnot equal to p. A picture follows.
PICT
Example B.1.5Consider A = B
(x,δ)
, an open ball in F^{p}. Then every point ofB
(x,δ)
is a limit point. (There are more general situations than ℝ^{p}in which thisassertion is false but these are of no concern in this book.)
so w_{k}→z. Furthermore, the w_{k} are distinct. Thus z is a limit point of A as claimed. This is
because every ball containing z contains infinitely many of the w_{k} and since they are all
distinct, they can’t all be equal to z.
A mapping f :
{k,k+ 1,k+ 2,⋅⋅⋅}
→ F^{p} is called a sequence. We usually write it
in the form
{aj}
where it is understood that a_{j}≡ f
(j)
. In the same way as for
sequences of real numbers, one can define what it means for convergence to take place.
Definition B.1.6A sequence,
{a }
k
is said toconverge to a if for every ε > 0
there exists n_{ε}such that if n > n_{ε}, then
|a − a |
n
< ε. The usual notation for this is
lim_{n→∞}a_{n} = a although it is often written as a_{n}→ a.
One can also define a subsequence in the same way as in the case of real valued sequences,
seen in calculus.
Definition B.1.7
{ank}
is a subsequenceof
{an}
if n_{1}< n_{2}<
⋅⋅⋅
.
Nothing changes if you use
∥⋅∥
_{∞} instead of
|⋅|
thanks to the equivalence of these
norms.The following theorem says the limit, if it exists, is unique. Thus the limit is well
defined.
Theorem B.1.8If a sequence,
{an}
converges to a and to b thena = b.
Proof:There exists n_{ε} such that if n > n_{ε} then
|an − a|
<
ε2
and if n > n_{ε}, then
|an − b|
<
ε2
. Then pick such an n.
ε -ε
|a− b| < |a− an|+ |an − b| < 2 + 2 = ε.
Since ε is arbitrary, this proves the theorem. ■
Then the following is about limit points.
Theorem B.1.9Let A be a nonempty set in F^{p}. A point a is a limit point ofA if and only if there exists a sequence of distinct points of A,
{ak}
which converges toa. 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 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 ≤ p, a_{k}∈ 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,
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. Thus at least one of them is not equal to a. 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.
■
Theorem B.1.10F^{p}is completely separable meaning that there exists acountable collection of open sets ℬ called a countable basis such that every open set isthe union of some subset of ℬ.
Proof: Let D consist of points d ∈ F^{p} such that each component is rational. Thus these
are points whose j^{th} component is of the form x + iy where x,y are both rational numbers.
This is a countable set by Theorem 3.2.7. Then consider ℬ to be the set of balls B
(d,r)
where
r ∈ ℚ and is positive and d ∈ D. Again this is a countable set by Theorem 3.2.7. It
suffices to show that every ball is the union of these sets. Let B
(x,R)
be a ball. Let
y ∈ B
(y,δ)
⊆ B
(x,R )
. Then there exists d ∈ B
(y,-δ)
10
. Let ε ∈ ℚ and
-δ
10
< ε <
δ
5
. Then
y ∈ B
(d,ε)
∈ℬ. Is B
(d,ε)
⊆ B
(x,R )
? If so, then the desired result follows because this
would show that every y ∈ B
(x,R)
is contained in one of these sets of ℬ which is contained
in B
. Therefore, every ball is the union of sets of ℬ and, since
every open set is the union of balls, it follows that every open set is the union of sets of ℬ.
■
Definition B.1.11Let S be a nonempty set. Then a set of open sets C iscalled an open cover of S if ∪C⊇S. (It covers up the set S. Think lilly pads coveringthe surface of a pond.)
Definition B.1.12The Lindeloff property says that whenever C is an opencover of a set S, there exists acountable subset of C denoted here by ℬ such that ℬ isalso an open cover of S.
Theorem B.1.13F^{p}has the Lindeloff property.
Proof: Let C be an open cover of a set S ⊆ F^{p}. Let ℬ be a countable basis. Such exists by
Theorem B.1.10. Let
ˆℬ
denote those sets of ℬ which are contained in some set of C. Thus
ˆℬ
is
a countable open cover of S. Now for B ∈ℬ, let U_{B} be a set of C which contains B. Letting
C^
denote these sets U_{B} it follows that
^C
is countable and is an open cover of S.
■
Definition B.1.14Let S be a nonempty set in F^{p}and let x ∈ F^{p}. Then the distanceof x to the set S is defined as
dist(x,S) ≡ inf{||x − y|| : y ∈ S}
The main result concerning this function is that it is Lipschitz continuous as described in
the following theorem.
Theorem B.1.15Let S≠∅ and consider f
(x)
≡dist
(x,S)
, then
|f (x)− f (ˆx)| ≤ ||x− y||
Here
||⋅||
is any norm on F^{p}, we have in mind either
||⋅||
_{∞}or
|⋅|
, the usual Euclidean normbut it will end up making no difference.
Proof: Say dist
(x,S)
≤dist
(ˆx,S)
. Otherwise, reverse the argument which follows. Then
for a suitable choice of y ∈ S,