2.1 Open And Closed Sets, Sequences, Limit Points, Completeness
It is most efficient to discus things in terms of abstract metric spaces to begin with.
Definition 2.1.1 A non empty set X is called a metric space if there is a
function d : X × X → [0,∞) which satisfies the following axioms.
- d =
≥ 0 and equals 0 if and only if x = y
- d +
This function d is called the metric. We often refer to it as the distance also.
Definition 2.1.2 An open ball, denoted as B
is defined as follows.
A set U is said to be open if whenever x ∈ U, it follows that there is r > 0 such that
⊆ U. More generally, a point x is said to be an interior point of U if there
exists such a ball. In words, an open set is one for which every point is an interior
For example, you could have X be a subset of ℝ and d
Then the first thing to show is the following.
Proposition 2.1.3 An open ball is an open set.
Proof: Suppose y ∈ B
We need to verify that y
is an interior point of B
= r − d
Then if z ∈ B
it follows that
Thus y ∈ B
Definition 2.1.4 Let S be a nonempty subset of a metric space. Then p is a
limit point (accumulation point) of S if for every r > 0 there exists a point different
than p in B
∩ S. Sometimes people denote the set of limit points as S′.
A related idea is the notion of the limit of a sequence. Recall that a sequence is really just
a mapping from ℕ to X. We write them as
if we want to emphasize
the values of n.
Then the following definition
is what it means for a sequence to
Definition 2.1.5 We say that x = limn→∞xn when for every ε > 0 there exists N
such that if n ≥ N, then
Often we write xn → x for short. This is equivalent to saying
Proposition 2.1.6 The limit is well defined. That is, if x,x′ are both limits of a
sequence, then x = x′.
Proof: From the definition, there exist N,N′ such that if n ≥ N, then d
and if n ≥ N′,
Then let M ≥
Let n > M.
Since ε is arbitrary, this shows that x = x′ because d
Next there is an important theorem about limit points and convergent sequences.
Theorem 2.1.7 Let S≠∅. Then p is a limit point of S if and only if there exists
a sequence of distinct points of S,
none of which equal p such that
is a limit point. Why does there exist the promissed convergent
sequence? Let x1 ∈ B
such that x1≠p.
have been chosen, let xn+1≠p
Then this constructs the
necessary convergent sequence.
⇐= Conversely, if such a sequence
exists, then for every
xn ∈ S
for all n
large enough. Hence, p
is a limit point because none of these xn
are equal to
Definition 2.1.8 A set H is closed means HC is open.
Note that this says that the complement of an open set is closed. If V is open, then the
complement of its complement is itself. Thus
an open set. Hence V C
Then the following theorem gives the relationship between closed sets and limit
Theorem 2.1.9 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 since H
is closed, its complement is open and so there
exists δ >
0 such that B
However, this prevents p
from being a limit
⇐= 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,
is open. Hence H
is closed. ■
Corollary 2.1.10 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
then from the definition of limit, it is a limit point of H
. Hence x ∈ H
⇐= Suppose the limit condition holds, why is H closed? Let x ∈ H′ the set of limit points
of H. By Theorem 2.1.7 there exists a sequence of points of H,
hn → x.
by assumption, x ∈ H.
contains all of its limit points and so it is closed by Theorem
Next is the important concept of a subsequence.
Definition 2.1.11 Let
n=1∞ be a sequence. Then if n1 < n2 <
is a strictly increasing sequence of indices, we say
k=1∞ is a subsequence of
The really important thing about subsequences is that they preserve convergence.
Theorem 2.1.12 Let
be a subsequence of a convergent sequence
xn → x. Then
Proof: Let ε > 0 be given. Then there exists N such that
It follows that if k ≥ N, then nk ≥ N and so
This is what it means to say limk→∞xnk = x. ■