B.1 Closed and Open Sets
The definition of open and closed sets is next.
Definition B.1.1 Let U be a set of points in Fp. A point, p ∈ U is said to be an
interior point of U 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. 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.
The first question to consider is whether an “open ball” is actually an open
Proposition B.1.2 An open ball in Fp is an open set.
Proof: Let z ∈ B
Is there a ball B
? Yes there is. Consider
is in this last ball, then
Thus every point of the ball is indeed an interior point and this shows the open ball is indeed
Note that the definition of open and closed does not depend on the choice of the norm
because of ∗ which implies that in ℝn, B
. 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
. However, notice that
a product of open intervals. This is especially convenient.
Theorem B.1.3 The 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 of
closed 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.
be some open sets and suppose
exists rk >
0 such that B
Let 0 < r ≤
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
is a finite set of closed sets. Then
is closed if its
complement is open. However,
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
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
Definition B.1.4 Let A be any nonempty set and let p be a point. Then p 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 p. A picture follows.
Example B.1.5 Consider A = B
, an open ball in Fp. Then every point of
is a limit point. (There are more general situations than ℝp in which this
assertion is false but these are of no concern in this book.)
If z ∈ B
for k ∈ ℕ
so wk → z. Furthermore, the wk are distinct. Thus z is a limit point of A as claimed. This is
because every ball containing z contains infinitely many of the wk and since they are all
distinct, they can’t all be equal to z.
A mapping f :
is called a sequence. We usually write it
in the form
where it is understood that
aj ≡ f
. In the same way as for
sequences of real numbers, one can define what it means for convergence to take place.
Definition B.1.6 A sequence,
is said to converge to a if for every ε >
there exists nε such that if n > nε, then
< ε. The usual notation for this is
= a although it is often written as an → a.
One can also define a subsequence in the same way as in the case of real valued sequences,
seen in calculus.
is a subsequence of
if n1 < n2 <
Nothing changes if you use
thanks to the equivalence of these
norms.The following theorem says the limit, if it exists, is unique. Thus the limit is well
Theorem B.1.8 If a sequence,
converges to a and to b then a
Proof: There exists nε such that if n > nε then
n > nε
. Then pick such an
Since ε is arbitrary, this proves the theorem. ■
Then the following is about limit points.
Theorem B.1.9 Let A be a nonempty set in Fp. 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. 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
such that a1≠a
Now supposing distinct points, a1,
have been chosen such that none are equal to a
for each k ≤ p, ak ∈ B
Then there exists an+1 ∈ B
Because of the definition of rn+1, an+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,
large enough. Thus B
contains infinitely many points
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 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
Therefore, if a ∈ AC,
since it is not a limit point of A,
there exists δ >
contains no points of
different than a.
itself is not in A
because a ∈ AC.
is entirely contained in
Since a ∈ AC
arbitrary, this shows every point of AC
is an interior point and so AC
Theorem B.1.10 Fp is completely separable meaning that there exists a
countable collection of open sets 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 ∈ Fp such that each component is rational. Thus these
are points whose jth 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
r ∈ ℚ
and is positive and d ∈ D
. Again this is a countable set by Theorem 3.2.7
suffices to show that every ball is the union of these sets. Let B
be a ball. Let
y ∈ B
Then there exists d ∈ B
Let ε ∈ ℚ
< ε <
y ∈ B
. Is B
? If so, then the desired result follows because this
would show that every
y ∈ B
is contained in one of these sets of
which is contained
is the union of sets of
. Let z ∈ B
. Therefore, every ball is the union of sets of
every open set is the union of balls, it follows that every open set is the union of sets of ℬ
Definition B.1.11 Let S be a nonempty set. Then a set of open sets C is
called an open cover of S if ∪C⊇S. (It covers up the set S. Think lilly pads covering
the surface of a pond.)
Definition B.1.12 The Lindeloff property says that whenever C is an open
cover of a set S, there exists a countable subset of C denoted here by ℬ such that ℬ is
also an open cover of S.
Theorem B.1.13 Fp has the Lindeloff property.
Proof: Let C be an open cover of a set S ⊆ Fp. Let ℬ be a countable basis. Such exists by
Theorem B.1.10. Let
denote those sets of
which are contained in some set of C
a countable open cover of
. Now for B ∈ℬ
, let UB
be a set of C
which contains B
denote these sets
it follows that
is countable and is an open cover of
Definition B.1.14 Let S be a nonempty set in Fp and let x ∈ Fp. Then the distance
of x to the set S is defined as
The main result concerning this function is that it is Lipschitz continuous as described in
the following theorem.
Theorem B.1.15 Let S≠∅ and consider f
is any norm on Fp, we have in mind either
, the usual Euclidean norm
but it will end up making no difference.
Proof: Say dist
. Otherwise, reverse the argument which follows. Then
for a suitable choice of
y ∈ S,
is arbitrary, this shows the claimed result. ■