11.1.1 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 11.1.1A non empty set X is called a metric space if there is a function d : X × X →
[0,∞) which satisfies the following axioms.
d
(x,y)
= d
(y,x)
d
(x,y)
≥ 0 and equals 0 if and only if x = y
d
(x,y)
+ d
(y,z)
≥ d
(x,z)
This function d is called the metric. We often refer to it as the distance.
Definition 11.1.2An open ball, denoted as B
(x,r)
is defined as follows.
B (x,r) ≡ {y : d(x,y) < r}
A set U is said to be open if whenever x ∈ U, it follows that there is r > 0 such that B
(x,r)
⊆ U. Moregenerally, a point x is said to be aninterior point of U if there exists such a ball. In words, an open set isone for which every point is an interior point.
For example, you could have X be a subset of ℝ and d
(x,y)
=
|x− y|
.
Then the first thing to show is the following.
Proposition 11.1.3An open ball is an open set.
Proof: Suppose y ∈ B
(x,r)
. We need to verify that y is an interior point of B
Definition 11.1.4Let 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
(p,r)
∩ 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
{xn }
or
{xn}
_{n=1}^{∞} if we want to emphasize the values of n. Then the
following definition is what it means for a sequence to converge.
Definition 11.1.5We say that x = lim_{n→∞}x_{n}when for every ε > 0 there exists N such that if n ≥ N,then
d(x,xn) < ε
Often we write x_{n}→ x for short. This is equivalent to saying
ln→im∞ d(x,xn) = 0.
Proposition 11.1.6The limit is well defined. That is, if x,x^{′}are both limits of asequence, thenx = x^{′}.
Proof:From the definition, there exist N,N^{′} such that if n ≥ N, then d
(x,xn)
< ε∕2 and if n ≥ N^{′},
then d
(x,xn)
< ε∕2. Then let M ≥ max
(N, N′)
. Let n > M. Then
′ ′ ε ε
d(x,x) ≤ d(x,xn)+ d(xn,x ) < 2 + 2 = ε
Since ε is arbitrary, this shows that x = x^{′} because d
(x,x′)
= 0. ■
Next there is an important theorem about limit points and convergent sequences.
Theorem 11.1.7Let S≠∅. Then p is a limit point of S if and only if there exists a sequence ofdistinct points of S,
{xn}
none of which equal p such that lim_{n→∞}x_{n} = p.
Proof:
=⇒
Suppose p is a limit point. Why does there exist the promised convergent sequence? Let
x_{1}∈ B
(p,1)
∩ S such that x_{1}≠p. If x_{1},
⋅⋅⋅
,x_{n} have been chosen, let x_{n+1}≠p be in B
(p,δn+1)
∩ S where
δ_{n+1} = min
{ }
n1+1,d(xi,p),i = 1,2,⋅⋅⋅,n
. Then this constructs the necessary convergent
sequence.
⇐= Conversely, if such a sequence
{xn }
exists, then for every r > 0, B
(p,r)
contains x_{n}∈ S for all n
large enough. Hence, p is a limit point because none of these x_{n} are equal to p. ■
Definition 11.1.8A set H is closed means H^{C}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
(VC)
^{C} = V an open set. Hence V^{C} is closed.
Then the following theorem gives the relationship between closed sets and limit points.
Theorem 11.1.9A 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 11.1.10A 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 from
the definition of limit, it is a limit point of H. Hence x ∈ H after all.
⇐= Suppose the limit condition holds, why is H closed? Let x ∈ H^{′} the set of limit points of H.
By Theorem 11.1.7 there exists a sequence of points of H,
{hn}
such that h_{n}→ x. Then by
assumption, x ∈ H. Thus H contains all of its limit points and so it is closed by Theorem 11.1.9.
■
Next is the important concept of a subsequence.
Definition 11.1.11Let
{xn}
_{n=1}^{∞}be a sequence. Then if n_{1}< n_{2}<
⋅⋅⋅
is a strictly increasingsequence of indices, we say
{xnk}
_{k=1}^{∞}is a subsequenceof
{xn}
_{n=1}^{∞}.
The really important thing about subsequences is that they preserve convergence.
Theorem 11.1.12Let
{xnk}
be a subsequence of a convergent sequence
{xn}
where x_{n}→ x.Then
lkim→∞ xnk = x
also.
Proof:Let ε > 0 be given. Then there exists N such that
d(xn,x) < ε if n ≥ N.
It follows that if k ≥ N, then n_{k}≥ N and so
d (xnk,x) < ε if k ≥ N.
This is what it means to say lim_{k→∞}x_{nk} = x. ■
Another useful idea is the distance to a set.
Definition 11.1.13Let
(X,d)
be a metric space and let S be a nonempty set in X. Then
dist(x,S ) ≡ inf {d (x,y) : y ∈ S} .
The following lemma is the fundamental result.
Lemma 11.1.14The function, x →dist
(x,S )
is continuous and in fact satisfies
|dist(x,S )− dist(y,S)| ≤ d(x,y).
Proof: Suppose dist
(x,S)
is as least as large as dist
(y,S )
. Then pick z ∈ S such that
d
(y,z)
≤dist
(y,S)
+ ε. Then
|dist(x,S)− dist(y,S )| = dist(x,S )− dist(y,S)
≤ d (x,z)− (d(y,z)− ε)
= d (x,z)− d (y,z)+ ε
≤ d (x,y)+ d (y,z)− d (y,z)+ ε
= d (x,y)+ ε.
Since ε > 0 is arbitrary, this proves the lemma. The argument is the same if dist