2.1 Closed and Open Sets
The definition of open and closed sets is next.
Definition 2.1.1 Let U be a set of points in Fp. A point, p ∈ U is said to be an interior point of U if
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
If you have two norms which are equivalent in the sense that
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
but from the above
1 < δ
. Now observe that
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 2.1.2 An open ball in Fp is an open set.
Proof: Let z ∈ B
Is there a ball B
? Yes there is. Consider
If y 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 open.
a product of open intervals. This is especially convenient. You should carefully write down the reasoning for
this from the definition of
Theorem 2.1.3 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
be some open sets and suppose
Then there exists rk >
0 such that
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, from DeMorgan’s laws, Problem 30
on Page 90
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 2.1.4 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 2.1.5 Consider A = B
, an open ball in Fp. Then every point of B
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
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 2.1.6 A sequence,
is said to converge to a if for every ε >
0 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
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 defined.
Theorem 2.1.8 If a sequence,
converges to a and to b then a
Proof: There exists nε such that if n > nε then
n > nε
pick such an
Since ε is arbitrary, this proves the theorem. ■
Then the following is about limit points.
Theorem 2.1.9 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.
Proof: Suppose first a is a limit point of A. There exists a1 ∈ B
such that a1≠a
supposing distinct points, a1,
have been chosen such that none are equal to a
and 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
it follows limm→∞am
. Conversely, if
there exists a sequence of distinct points of A
converging to a,
large enough. Thus B
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 2.1.10 A set H is closed if and only if it contains all of its limit points.
be closed and let p
be a limit point. We need to verify that p ∈ H
. If it is not, then
is closed, its complement is open and so there exists δ >
0 such that B
this prevents p
from being a limit point.
⇐= Next suppose H has all of its limit points. Why is HC open? If p ∈ HC then it is not a limit point
and so there exists δ > 0 such that B
has no points of
. In other words, HC
is open. Hence H
Corollary 2.1.11 A set H is closed if and only if whenever
is a sequence of points of H
which converges to a point x, it follows that x ∈ H.
is closed and hn → x.
If x ∈ H
there is nothing left to show. If x
x ∈ B
for some r >
0. If ε < r,
then hn ∈ B
large enough but hn≠x
assumption. Hence x
is a limit point of H
and so x ∈ H
⇐= Suppose the limit condition holds. Why is H closed? In other words, why is HC open? If HC is not
open, then there exists x
which is not an interior point of HC
. Thus B
contains a point
pn ∈ H
. However, this requires that pn → x
and so x ∈ H
by the limit condition. This would be a
contradiction. Thus HC
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
It is only the properties of a norm which are needed.
Theorem 2.1.12 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 1.2.7. 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 1.2.7
. It suffices to show that every ball is the union of these sets.
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
contained in B
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 2.1.13 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 2.1.14 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 2.1.15 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
denote those sets of
which are contained in some set of C
is a countable open cover
. 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 2.1.16 Let S be a nonempty set in Fp and let x ∈ Fp. Then the distance of x to the set S is
The main result concerning this function is that it is Lipschitz continuous as described in the following
Theorem 2.1.17 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
y ∈ S,
is arbitrary, this shows the claimed result. ■