The Bolzano Weierstrass Theorem for Sets and
Set Ideas
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
February 1, 2017
Outline
Uniqueness of Limits And So Forth
The Bolzano Weierstrass Theorem For Sets
Set Ideas
Theorem
If the sequence (xn ) converges, then the limit is unique.
Proof
Assume xn → a and xn → b. Then for an arbitrary > 0, there is an
N1 and an N2 so that
n > N1
⇒ |xn − a| < /2
n > N2
⇒ |xn − b| < /2
Now pick a P > max{N1 , N2 }. Then, we have
|a − b| =
|a − xP + xP − b| ≤ |a − xP | + |xP − b| < /2 + /2 = .
Since > 0 is arbitrary and |a − b| < for all > 0, this shows a = b.
Theorem
Assume (xn ) converges and (xnk ) is a subsequence. Then (xnk )
converges to the same limiting value.
Proof
Since xn → a for some a, for arbitrary > 0, there is an N so that
n > N ⇒ |xn − a| < . In particular, nk > N ⇒ |xnk − a| < . This
says xnk → a also.
Theorem
Bolzano Weierstrass Theorem For Sets
Every bounded infinite set of real numbers has at least one
accumulation point.
Proof
We let the bounded infinite set of real numbers be S. We know there
is a positive number B so that −B ≤ x ≤ B for all x in S because S is
bounded.
Step 1:
By a process essentially the same as the subdivision process in the
proof of the Bolzano Weierstrass Theorem for Sequences, we can find
a sequence of sets
J0 , J1 , . . . , Jp , . . .
These sets have many properties.
Proof
I
`p = B/2p−1 for p ≥ 1 ( remember B is the bound for S )
I
Jp ⊆ Jp−1 ,
I
Jp = [αp , βp ] and Jp contains IMPs of S.
I
−B = α0 ≤ . . . ≤ αp < βp ≤ . . . ≤ β0 = B
Since J0 contains IMPs of S choose x0 in S from J0 . Next, since J1
contains IMPs of S, choose x1 different from x0 from S. Continuing (
note we are using what we called our relaxed use of POMI here!) we see
by an induction argument we can choose xp in S, different from the
previous ones, with xp in Jp .
We also know the sequence (αp ) has a finite supremum α and the
sequence (βp ) has a finite infimum β. Further, we know αp ≤ α, β ≤ βp
for all p. Since the lengths of the intervals Jp go to 0 as p → ∞, we also
know α = β and the sequence (xp ) converges to α.
Proof
It remains to show that α is an accumulation point of S. Choose any
r > 0. Since `p = B/2p−1 , we can find an integer P so that B/2P−1 < r .
Thus, since the ball Br (α) = (α − r , α + r ) ( an open interval of the real
line ), we see by our construction process that α ∈ JP and
xp ∈ JP = [αP , βP ] for any p ≤ P. So we know also
|xp − α|
≤ |βp − αp | ≤ |βP − αP | =
B
<r
2P−1
Hence, xp is in Br (α) for all p ≥ P and so Br (α) contains IMPs of S
distinct from α. Since our choice of r > 0 was arbitrary, this tells us α is
an accumulation point of S.
Comment
Note the Bolzano Weiesrstrass Theorem for Sets shows the bounded
infinite set S must have at least one accumulation point α and we found
out much more:
∃ a sequence (xp ) ⊆ S so that xp → α
Comment
Note we don’t know α ∈ S.
This leads to a new definition.
Definition
Let S be a nonempty set of real numbers. We say the number a is a
cluster point of S if there is a sequence (xp ) ⊆ S with each xp 6= a so
that xp → a. Note any accumulation point is also a cluster point and
any cluster point is an accumulation point.
We can therefore restate our theorem like this:
Theorem
Bolzano Weierstrass Theorem for Sets
Every bounded infinite set of real numbers has at least one cluster
point.
Proof
The proof is the same: we just note the accumulation point we have
found is also a cluster point.
Example
S = {2}. This is a set of one element only. Note 2 is not an
accumulation point of S as B0.5 (2) = (1.5, 2.5) only contains 2 from S.
Since no other points of S are in this circle, 2 can not be an accumulation
point of S. It is also not a cluster point because we can’t find a sequence
(xn ) in S with each xn 6= 2 that converges to 2. The only sequence that
works would be the constant sequence (xn = 2) which we can’t use.
Example
S = (0, 4). If c is in S, the sequence (xn = c + 1/n) for n small enough
will be in S and will converge to c. So every c in (0, 4) is a cluster point.
We note 0 is also a cluster point because the sequence (xn ) = (1/n) is
contained in S and converges to 0. We also see 4 is also a cluster point
because the sequence (xn ) = (4 − 1/n) is contained in S and converges
to 4. The set of cluster points of S is thus [0, 4].
Example
S = {1, 2, 3, 4} ∪ (6, 7] ∪ {10}. What are the cluster points of S? You
should be able to see that the cluster points of S are [6, 7] and the
accumulation points of S are also [6, 7]
Now it is time to define more types of sets of numbers: open sets,
closed sets and so on. These are useful concepts we will use a lot.
Definition
Let S be a set of real numbers. We say S is an open set for every
point p in S. we can find a ball or radius r centered at p which is
strictly contained in S. That is ∀ p ∈ S ∃ r > 0 3 Br (p) ⊂ S.
Example
S = {2} is not an open set because every circle Br (2) contains lots of
points not in S!
Example
S = (0, 4) is an open set because for every p in S, we can find an r > 0
with Br (p) ⊂ (0, 4).
Example
S = (0, 5] is not an open set because although for every p in (0, 5), we
can find an r > 0 with Br (p) ⊂ (0, 5], we can not do that for 5. Every
circle centered at 5 spills outside of (0, 5].
We can define more types of sets.
Definition
S is a set of real numbers. Note 6∈ means not in . Then, we define
S C = {x : x 6∈ S}, the complement of S. We say S is closed if its
complement S C is open.
Definition
S is a set of real numbers. Then, we define p in S to be an interior
point of S if we can find a positive r so that the ball Br (p) ⊂ S. Note
from our definition of open set, every point in an open set is an
interior point. Then we have
Int S
=
{x : x is an interior point of S},
∂S
=
{x : all Br (x) contains points of both S and S C },
S0
=
S ∪ ∂S,
the interior of S
the boundary of S
the closure of S
Note if S is open, Int(S) = S. Also a point x which is in ∂S is called
a boundary point. In the set S = {2, 3, 4}, 2, 3 and 4 are boundary
points but they are not accumulation points as each Br (x) only
contains x.
Another interesting type of point is a limit point.
Definition
Let S be a set of numbers. We say p is a limit point of S if there is a
sequence of points in S, (xn ), so that xn → p.
Note that an isolated point, like 2 ∈ S = {2, 3} ∪ (4, 5) is a limit point
since we can choose the constant sequence (xn = 2) and then xn → 2
even though 2 is not a cluster point nor is it an accumulation point.
We can use this idea to characterize a closed set nicely.
Theorem
A set S is closed if and only if it contains all its limit points.
Proof
(=⇒):
If S is closed, then its complement is open. Assume (xn ) is a sequence in
S with limit point x and x 6∈ S, i.e. x ∈ S C . Then x is an interior point
of S C and so there is a radius r > 0 so that Br (x) ⊂ S C . Since xn → x,
given = r /2, there is an N so that |x − xn | < r /2 when n > N. This
says xn ∈ Br (x) ⊂ S C . But xn ∈ S so this is a contradiction. So each
limit point of S must be in S.
(⇐=):
We assume all of S’s limit points are in S. Let x ∈ S C . If we could not
find a radius r > 0 so that Br (x) ⊂ S C , then SC would not be open and
so S would not be closed. But if this were the case, for each rn = 1/n,
there is an xn ∈ S with |x − xn | < 1/n. But this says xn → x with
(xn ) ⊂ S. So x is a limit point of S and by assumption must be in S.
This is a contradiction and so S C must be open implying S is closed.
Example
If S = (0, 2), 0 and 2 are accumulation points and they are also boundary
points.
If S = [0, 2], 0 and 2 are accumulation points and they are also boundary
points.
Example
If S = {2} ∪ (3, 4),
Int S = (3, 4)
S C = (−∞, 2) ∪ (2, 3] ∪ [4, ∞),
∂S = {2} ∪ {3} ∪ {4}
S 0 = {2} ∪ [3, 4],
(S 0 )C = (−∞, 2) ∪ (2, 3) ∪ (4, ∞) which is open and so S 0 is closed.
Example
Let S = (2/n + 3)n≥1 . Then S = {5, 4, 3 23 , 3 24 , . . .}.
IntS = ∅; there are no interior points.
SC
=
(−∞, 3]
2
2 2
∪(4, 5) ∪ (3 , 4) ∪ (3 , 3 ), . . .
3
4 3
∪(5, ∞)
which is not an open set because 3 is a boundary point and so it is not
an interior point.
∂S = {5, 4, 3 23 , 3 24 , . . .} and {3}. Note 3 is both an accumulation point
and a boundary point. But each entry in the sequence, i.e. 3 32 , is not an
accumulation point although it is a boundary point.
S 0 = {5, 4, 3 23 , 3 24 , . . .} and {3}.
Homework 9
9.1 Let S = {2, 3, 6}. Find Int(S), ∂S, S 0 and the accumulation points
of S.
9.2 Let S = (2 + 1/n)n≥1 . Find Int(S), ∂S, S 0 and the accumulation
points of S.
9.3 Let P(x) = b0 + b1 x + b2 x 2 + . . . + bn x n with bn 6= 0 be an
arbitrary polynomial of degree n. If an → a, prove P(an ) → P(a)
using the POMI. Here the basis is an arbitrary polynomial of degree
1, i.e. P(x) = b0 + b1 x with b1 6= 0 and our algebra of sequence
limits results show the proposition is true here. For the inductive
step, assume it is true for an abitrary polynomial of degree k,
P(x) = b0 + b1 x + . . . + bk x k and then note a polynomial of degree
k + 1 looks a polynomial of degree ≤ k plus the last term bk+1 x k+1 .
You should be able to see how to finish the argument from that.