Adherent Points, Accumulation Points, Isolated Points
Definition
Suppose that (X , d) is a metric space and A is a subset of X .
Functional Analysis
(a) A point x ∈ X is called an adherent point of A if every ball Br (x)
contains at least one point y ∈ A.
(b) An adherent point x ∈ X of A is called an accumulation point of A
if every ball Br (x) contains at least one point y ∈ A such that y 6= x.
The set of accumulation points of A is called the derived set and is
denoted F 0 .
(c) An adherent point of A, which is not an accumulation point of A, is
called an isolated point of A.
Lecture 2: Metric Spaces – Accumulation Points, Closure,
Dense Subsets, Separable Spaces
Bengt Ove Turesson
September 2, 2016
Remark
Notice that any adherent point of A is either an accumulation point
or an isolated point of A.
Notice also that if x is an isolated point of A, then x ∈ A and there
exists a ball Br (x) that contains no other points from A than x.
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The Closure
The Closure
Definition
Proposition
Suppose that (X , d) is a metric space and A is a subset of X . The
closure A of A is the set of all adherent points of A.
Suppose that (X , d) is a metric space and A is a subset of X . Then
the closure A of A is a closed subset of X .
Remark
Proof.
c
Notice that A ⊂ A.
We need to show that A is open.
c
If A = ∅, there is nothing to prove, so let us assume that
c
A 6= ∅.
c
Take x ∈ A and choose r > 0 such that Br (x) ∩ A = ∅.
If y ∈ Br (x), then since Br (x) is open, there exists a number
s > 0 such that Bs (y ) ⊂ Br (x).
c
It follows that Bs (y ) ∩ A = ∅, so y ∈ A .
c
This shows that Br (x) ⊂ A .
c
Hence, A is open.
Example
We claim that (a, b) = [a, b] for −∞ < a < b < ∞.
Indeed, every point in (a, b) is an accumulation point of (a, b).
Furthermore, the end-points a and b of (a, b) are
accumulation points of (a, b).
Since (a, b) has no isolated points, the claim follows.
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The Closure
The Closure
Proposition
Proposition
Suppose that (X , d) is a metric space. A subset F of X is closed if
and only if F = F .
Suppose that (X , d) is a metric space and A is a subset of X .
Then the closure A of A is the smallest closed superset of A with
respect to inclusion.
Proof.
Proof.
Sufficiency: This part is obvious since F is closed.
Necessity: Suppose that F is closed, which means that F c is
open. Then, for any x ∈ F c there exists a number r > 0 such
c
that Br (x) ⊂ F c . It follows that F c ⊂ F , and hence that
F ⊂ F . Since F ⊂ F for any set F , this shows that
F = F.
We already know that A ⊂ A and that A is closed.
Suppose that F is a closed subset of X such that A ⊂ F .
It then follows that A ⊂ F .
This finishes the proof since F = F .
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Dense Subsets of a Metric Space
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Separable Metric Spaces
Definition
Suppose that (X , d) is a metric space and A is a subset of X .
Then A said to be dense in X if A = X .
Definition
Example
A metric space (X , d) is said to be separable if it contains a
countable, dense subset.
It is well-known that Q is dense in R: Given a real number,
any open interval that contains the number, also contains a
rational number not equal to the given number.
Example
The metric spaces Rd and Cd are separable. This follows from the
previous example and the fact that Q is countable.
This fact implies that the set of rational complex numbers,
i.e., complex numbers with rational real and imaginary parts,
is dense in C.
From this it follows that the set of vectors with rational
entries is dense in Rd and that the set of vectors with rational
complex entries is dense in Cd .
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Separable Metric Spaces
Separable Metric Spaces
Example
Example
Let us show that `p is separable for 1 ≤ p < ∞.
We shall show that the space `∞ is not separable.
Let E be the set of sequences (yj )∞
j=1 , where every yj is a rational
complex number and not more than a finite number of yj are nonzero.
Then E is countable (exercise).
It suffices to show that if x ∈ `p , then x is an accumulation point of E .
For an arbitrary ε > 0, first choose N so large that
P
∞
p
p
j=N+1 |xj | < ε /2.
For j = 1, . . . , N then choose rational complex numbers yj such that
yj 6= xj and |xj − yj | < ε/(2N)1/p .
Finally set yj = 0 for j = N + 1, N + 2, . . . and let y = (yj )∞
j=1 .
Then
N
∞
X
X
dp (x, y )p =
|xj − yj |p +
|xj |p < εp ,
j=1
Let Y be the subspace of `∞ that consists of all sequences,
where all elements in the sequence are 0 or 1.
Then Y is uncountable since every element in Y corresponds
to a number in [0, 1] written in the basis 2 and [0, 1] is
uncountable.
The distance between two different elements of Y is 1, which
implies that the balls B1/2 (x), where x ∈ Y , are disjoint.
Suppose that {x1 , x2 , . . .} were a countable, dense subset
of `∞ .
Then every ball B1/2 (x) would contain at least one xn and no
two balls could contain the same xn . But this is impossible
since the set of such balls is uncountable.
j=N+1
so dp (x, y ) < ε.
This shows that x is an accumulation point of E .
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Separable Metric Spaces
Example
According to Weierstrass’ approximation theorem, C ([a, b]) is
separable: The countable set of polynomials with rational
coefficients is dense in C ([a, b]).
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