Definition 2.1.1Let U be a set of points in F^{p}. A point, p ∈ U is said to be an interior point of U ifwhenever
∥x− p∥
is sufficiently small, it follows x ∈ U also. The set of points x which are closer to p thanδ 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 of U if thereexists δ > 0 such that p ∈ B
(p,δ)
⊆ U. An open set is one for which every point of the set is an interiorpoint.Closed sets are those which arecomplements of open sets. Thus H is closed means H^{C}isopen.
If you have two norms which are equivalent in the sense that
δ∥x∥ ≤ ∥x∥1 ≤ Δ ∥x∥
then a set U is open with respect to one norm if and only if it is open with respect to the other. Indeed, if
x ∈ U which is open with respect to
∥⋅∥
, then there is B_{∥⋅∥
}
(x,δ)
⊆ U but from the above
inequality,
( 1 )
B∥⋅∥1 x,-δ ⊆ B ∥⋅∥(x,δ) ⊆ U
Δ
because if
∥x − y∥
_{1}<
δ-
Δ
, then
∥x − y∥
≤ Δ
∥x − y∥
_{1}< δ. Now observe that
1- 1
Δ ∥x∥1 ≤ ∥x∥ ≤ δ ∥x∥1
You should write down the reasoning to this carefully.
The first question to consider is whether an “open ball” is actually an open set.
Proposition 2.1.2An 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.
■
Notice that
∏n
B ∞ (p,r) = (pi − r,pi + r)
i=1
a product of open intervals. This is especially convenient. You should carefully write down the reasoning for
this from the definition of
∥⋅∥
_{∞}.
Theorem 2.1.3The intersection of any finite collection of open sets is open.The union of anycollection of open sets is open.The intersection of any collection of closed setsis closed and theunion of any finite collection of closed setsis 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
{U ,⋅⋅⋅,U }
1 m
be some open sets and suppose p ∈∩_{k=1}^{m}U_{k}. Then there exists r_{k}> 0 such that
B
(p,r )
k
⊆ U_{k}. Let 0 < r ≤ min
(r ,r ,⋅⋅⋅,r )
1 2 m
. 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
{H ,⋅⋅⋅,H }
1 m
is a finite set of closed sets. Then ∪_{k=1}^{m}H_{k} is closed if its complement is open.
However, from DeMorgan’s laws, Problem 30 on Page 90,
(∪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 2.1.4Let A be any nonempty set and let p be a point. Then p is said to be a limit pointof A if for every r > 0,B
(p,r)
contains a point of A which is not equal to p. A picture follows.
PICT
Example 2.1.5Consider A = B
(x,δ)
, an open ball in F^{p}. Then every point of B
(x,δ)
is a limitpoint. (There are more general situations than ℝ^{p}in which this assertion is false but these are of noconcern 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 2.1.6A sequence,
{ak}
is said toconverge to a if for every ε > 0 there exists n_{ε}suchthat if n > n_{ε}, then
|a− an|
< ε. The usual notation for this is lim_{n→∞}a_{n} = a although it is oftenwritten 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 2.1.7
{an }
k
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 2.1.8If 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|
<
ε
2
and if n > n_{ε}, then
|an − b|
<
ε
2
. Then
pick such an n.
|a− b| < |a − an |+|an − b | < ε + ε= ε.
2 2
Since ε is arbitrary, this proves the theorem. ■
Then the following is about limit points.
Theorem 2.1.9Let A be a nonempty set in F^{p}. A point a is a limit point of A if and only if thereexists a sequence of distinct points of A,
{ak}
which converges to a.
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,1∕k)
, let
{ --1-- }
0 < rn+1 < min n + 1,∥a− a1∥,⋅⋅⋅,∥a − an∥ .
Then there exists a_{n+1}∈ B
(a,r )
n+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 − a ∥
m
< 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.
■
Theorem 2.1.10A set H is closed if and only if it contains all of itslimit points.
Proof:
=⇒
Let H be closed and let p be a limit point. We need to verify that p ∈ H. If it is not, then
since H is closed, its complement is open and so there exists δ > 0 such that B
(p,δ)
∩ H = ∅. However,
this prevents p from being a limit point.
⇐= Next suppose H has all of its limit points. Why is H^{C} open? If p ∈ H^{C} then it is not a limit point
and so there exists δ > 0 such that B
(p,δ)
has no points of H. In other words, H^{C} is open. Hence H is
closed. ■
Corollary 2.1.11A set H is closed if and only if whenever
{hn}
is a sequence of points of Hwhich converges to a point x, it follows that x ∈ H.
Proof:
=⇒
Suppose H is closed and h_{n}→ x. If x ∈ H there is nothing left to show. If x
∈∕
H, then
x ∈ B
(x,r)
⊆ H^{C} for some r > 0. If ε < r, then h_{n}∈ B
(x,ε)
for all n large enough but h_{n}≠x by
assumption. Hence x is a limit point of H and so x ∈ H after all.
⇐= Suppose the limit condition holds. Why is H closed? In other words, why is H^{C} open? If H^{C} is not
open, then there exists x
∕∈
H which is not an interior point of H^{C}. Thus B
(x,1∕n)
contains a point
p_{n}∈ H. However, this requires that p_{n}→ x and so x ∈ H by the limit condition. This would be a
contradiction. Thus H^{C} is open as claimed. ■
Note that the above theorems and corollary apply just as well to any normed vector space. This is a
vector space which has a norm satisfying
∥x∥ ≥ 0 and ∥x∥ = 0 if and only if x = 0 (2.8)
(2.8)
For α a scalar, ∥αx ∥ = |α|∥x∥ (2.9)
(2.9)
∥x + y∥ ≤ ∥x∥ + ∥y∥ (2.10)
(2.10)
It is only the properties of a norm which are needed.
Theorem 2.1.12F^{p}is completely separablemeaning that there exists a countablecollection of opensets ℬ called a countable basis such that every open set is the 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 1.2.7. 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 1.2.7. 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
( δ)
y,10
. Let ε ∈ ℚ and
1δ0
< ε <
δ5
. 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 2.1.13Let S be a nonempty set. Then a set of open sets C is called an open cover ofS if ∪C⊇S. (It covers up the set S. Think lilly pads covering the surface of a pond.)
Definition 2.1.14The Lindeloff property says that whenever C is an open cover of a set S, thereexists acountable subset of C denoted here by ℬ such that ℬ is also an open cover of S.
Theorem 2.1.15F^{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
2.1.12. 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 2.1.16Let S be a nonempty set in F^{p}and let x ∈ F^{p}. Then the distance of x to the set S isdefined 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 2.1.17Let S≠∅ and consider f
(x)
≡dist
(x,S)
, then
|f (x )− f (ˆx)| ≤ ∥x− ˆx∥
Here
∥⋅∥
is any norm on F^{p}, we have in mind either
∥⋅∥
_{∞}or
|⋅|
, the usual Euclidean norm but it will endup 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,