CHAPTER 3
2. Open and closed sets
Definition 3.2.1. An -ball around the point a ∈ R is the set
B(a) = {x ∈ R : |x− a| < } = (a− , a+ ).
The book uses the symbol V, and calls it a neighborhood. (Typically neighborhood can
be any open set, around a point, not simply the interval.)
Definition 3.2.2. A set A ⊂ R is open if for every point a ∈ A, there is some
> 0 such that B(a) ⊂ A.
Which of the following sets are open?
(1) The empty set.
This set is open. It is vacuously true that every point of ∅ contains an
-ball surrounding it that is contained in ∅.
(2) R.
This set is open. If a ∈ R, then B1(a) ⊆ R.
(3) R \ {0}.
This set is open. If a ∈ R \ {0}, then B|a|(a) ⊆ R \ {0}.
(4) {0}.
This set is not open. For all > 0, B(0) 6⊆ {0}.
(5) Q.
This set is not open since for all > 0, B(0) contains irrational numbers.
1
2 3
(6) (0, 1).
This set is open since for all a ∈ (0, 1), Bmin(a,1−a)(a) ⊆ (0, 1).
(7) [0, 1].
This set is not open since for all > 0, B(0) contains negative elements.
(8) [0, 1).
This set is not open for the same reason as above.
(9) Find a non-empty, bounded open set that is not an open interval.
(0, 1) ∪ (1, 2).
(10) True/False: The union of two open sets is an open set.
True. If A and B are open and x ∈ A ∪ B, then either x ∈ A, in which
case there is some > 0 so that B(x) ⊆ A ⊆ A∪B, or x ∈ B, in which case
again there is some so B(x) ⊆ B ⊆ A ∪B.
(11) True/False: The union finitely many open sets is an open set.
True. This works similarly to the above.
(12) True/False: The union of infinitely many open sets is an open set.
True. This works similarly to the above.
(13) True/False: The intersection of two open sets is an open set.
True. If A and B are open sets and x ∈ A ∩ B, then there is some
B(x) ⊆ A and some B′(x) ⊆ B. Then Bmin(,′)(x) ⊆ A ∩B.
(14) True/False: The intersection of finitely many open sets is an open set.
True. The minimum of finitely many positive real numbers is still positive.
2. OPEN AND CLOSED SETS 3
(15) True/False: The intersection of infinitely many open sets is an open set.
False. Let An = (−1/n, 1/n). Then
⋂∞
n=1An = {0}, which is not open.
Definition 3.2.3. A point x ∈ R is a limit point of a set A if there exists a
sequence {an}∞n=1 contained in A with lim an = x and an 6= x for any n. The set of
limit points of A is denoted L(A).
Definition 3.2.4. A set is closed if it contains all of its limit points.
Theorem 3.2.5 (3.2.13). If A ⊆ R, then A is closed if and only if Ac is open.
Definition 3.2.6. Given any set A, the closure A¯ is a closed set formed by the
union of A and all limit points of A; ie A¯ = A ∪ L(A).
Definition 3.2.7. A point x in a set A is an isolated point if it is an element
of A and is not a limit point of A. The set of isolated points of A is denoted I(A).
Thus A¯ = I(A) ∪ L(A).
Determine which sets are closed.
(1) The empty set.
It is closed. There is no point in ∅ that is not a limit point.
(2) R.
It is closed. If (an) is a convergent real sequence, then limn→∞ an ∈ R.
(3) R \ {0}.
It is not closed because limn→∞ 1n = 0 and
1
n
∈ R \ {0} for all n ≥ 1.
(4) {0}.
It is closed. The set has no limit points.
(5) Q.
It is not closed. We have that
√
2 ∈ L(Q) since √2 = limn→∞ bn
√
2c
n
.
(6) (0, 1).
4 3
It is not closed since 0 = limn→∞ 1n is a limit point of (0, 1).
(7) [0, 1].
It is closed. If x is a limit point of [0, 1], then x = limn→∞ xn where
xn ∈ [0, 1], since 0 ≤ xn ≤ 1, by the order limit theorem, 0 ≤ x ≤ 1.
(8) [0, 1).
It is not closed since 1 is a limit point.
(9) Find a non-empty closed set A which contains no limit points.
{0}.
(10) Find a non-empty closed set A which contains no isolated points.
[0, 1].
(11) Find a non-empty closed set A which contains both limit points and isolated
points.
{−1} ∪ [0, 1].
(12) Consider A = { 1
n
: n ∈ N}. Find L(A) and I(A).
L(A) = {0}, I(A) = A.
(13) Find a non-empty, bounded closed set that is not a closed interval.
[0, 1] ∪ [2, 3].
(14) True/False: The union of two closed sets is a closed set.
True. If A and B are closed, then Ac and Bc are open. Thus, Ac ∩Bc is
open, but this equals (A ∪B)c.
(15) True/False: The union finitely many closed sets is a closed set.
2. OPEN AND CLOSED SETS 5
True. The same argument above shows that if A1, . . . , An are closed, then
(A1 ∪ A2 ∪ · · · ∪ An)c = Ac1 ∩ Ac2 ∩ · · · ∩ Acn
is a finite intersection of open sets and is hence open.
(16) True/False: The union of infinitely many closed sets is a closed set.
False.
⋃∞
n=1
[
0, 1− 1
n
]
= [0, 1) is not closed.
(17) True/False: The intersection of two closed sets is a closed set.
True. If A1 and A2 are open, then (A1 ∩ A2)c = Ac1 ∪ Ac2 is open and so
A1 ∩ A2 is closed.
(18) True/False: The intersection of finitely many closed sets is a closed set.
True. The same argument as above works.
(19) True/False: The intersection of countably many closed sets is a closed set.
True. The same argument as above works since an arbitrary union of
open sets is open.
(20) True/False: If a is a limit point of A, then a ∈ A.
False. This is only true if A is closed.
(21) True/False: If a is a limit point of A, then for every > 0, the -ball around
a, denoted by B(a), and defined as the interval (a − , a + ), intersects A
in a point other than a.
True. If (an) is a sequence of elements in A converging to a with an 6= a,
then for all > 0, ∃N ∈ N so that n ≥ N implies that |an − a| < . This
means that A ∩B(a) contains {aN , aN+1, . . .}.
(22) True/False: If for every > 0, the -ball around a intersects A in a point
other than a, then a is a limit point of A.
True. Let an be an element of B 1
n
(a)∩A other than a. Then limn→∞ an =
a and so a is a limit point of A.
6 3
(23) True/False: If a set is closed, then it contains all of its limit points.
True. This is the definition.
(24) True/False: If a set is open, then it contains all of its limit points.
False. (0, 1) doens’t contain 0 or 1.
(25) True/False: If a set contains all of its limit points, then it is closed.
True. This is the definition.
(26) True/False: If a set contains all of its limit points, then it is open.
False. [0, 1] contains all its limit points and isn’t open.
(27) True/False: If a set contains all of its limit points, then it is not open.
False. R contains all of its limits points and it is open.
Q: Which subsets of R are both open and closed?
3. COMPACT SETS 7
3. Compact sets
Definition 3.3.1. A set A ⊂ R is compact if every sequence in A contains a
convergent subsequence, which converges to a limit in A.
Determine which sets are compact.
(1) The empty set.
This set is compact. There isn’t a sequence of elements in ∅ that doesn’t
have a convergent subsequence.
(2) R.
This set is not compact. If an = n, then there is no convergent subse-
quence.
(3) R \ {0}.
This set is not compact. If an = 1/n, then there is no subsequence (ank)
that converges to an element of R \ {0}.
(4) {0}.
This set is compact. The only sequence is an = 0.
(5) Q.
This set is not compact. Let an =
bn√2c
n
. Then each term is in Q, but
limn→∞ an =
√
2 6∈ Q.
(6) (0, 1).
This set is not compact. Let an = 1/n. Then an → 0 6∈ (0, 1).
(7) [0, 1].
This set is compact. If (an) ⊆ [0, 1] is any sequence, by BW, there is
a convergent subsequence (bn). We have 0 ≤ bn ≤ 1, and the order limit
theorem gives that 0 ≤ limn→∞ bn ≤ 1. This the limit is contained in [0, 1].
8 3
(8) [0, 1).
Not compact. Let an = 1− 1/n.
(9) {0} ∪ {1}.
Compact. If (an) is any sequence with an ∈ {0} ∪ {1}, then either there
are infinitely many n with an = 0 or infinitely many n with an = 1.
Lemma 3.3.2. If K is compact, then min{x ∈ K} and max{x ∈ K} exist.
Claim: If K is compact, then it is bounded.
Proof of claim: If K is unbounded, then for all n ∈ N, ∃an ∈ K so that |an| ≥ n.
The sequence (an) has the property that every subsequence is unbounded, and so no
subsequence can converge.
Claim: If K is compact, sup(K) ∈ K.
We have that for all > 0, ∃k ∈ K so that sup(K) − k < . For each n ∈ N let
an ∈ K be an element so that 0 ≤ sup(K) − an < 1n . Then limn→∞ an = sup(K).
Since K is compact, there is a subsequence (ank) that converges to an element of K,
but every subsequence converges to sup(K). Thus sup(K) = limk→∞ ank ∈ K.
A similar argument proves that inf(K) ∈ K.
3. COMPACT SETS 9
Theorem 3.3.3 (Heine-Borel). A set K ⊂ R is compact if and only if it is both
closed and bounded.
We showed on the previous page that if K is compact, then K is bounded. Claim:
If K is compact, then K is closed.
Proof of claim: Suppose that x is limit point of K. Then there is a sequence
(an) so that an ∈ K for all n and limn→∞ an = x. The assumption that K is
compact means there is a subsequence (ank) that converges to an element of K.
Thus, x = limk→∞ ank ∈ K.
So if K is compact, then K is closed and bounded.
Suppose that K is a closed and bounded set. Suppose that (an) is a sequence of
elements in K. Since K is bounded, there is some M ∈ R so that |an| ≤ M for all
n. By the BW theorem, there is a subsequence (ank) so that limk→∞ ank exists. Let
x = limk→∞ ank . If x = ank for some k, then x ∈ K. If x 6= ank for all k, then x
is by definition a limit point of K. This implies that x ∈ K. This proves that K is
compact.
10 3
Theorem 3.3.4. If Kn, n = 1, 2, . . . , is a sequence of nonempty, compact and
nested sets (Kn+1 ⊂ Kn for all n), then ∩∞n=1Kn 6= ∅.
Let x1, x2, . . ., be a sequence of elements so that xn ∈ Kn for all n ≥ 1. Since
(xn) ⊆ K1 and K1 is compact, there is a subsequence (xnk) that converges to an
element of K1. Let x = limk→∞ xnk . I claim that x ∈
⋂∞
n=1Kn.
If x = xnk for some k ≥ n, then x ∈ Knn ⊂ Kn. If x 6= xnk for no values of k ≥ n,
then x is a limit point of Kn. Since Kn is closed, x ∈ Kn.
3. COMPACT SETS 11
Definition 3.3.5. Let A ⊂ R. A open cover for A is a collection of open sets
{Aα : α ∈ I}, whose union contains A. Given an open cover for A, a finite subcover
is a finite sub-collection of open sets from the original open cover whose union also
contains A.
Examples:
(1) Consider R. An example of an open cover is the collection of overlapping
intervals, for each n ∈ Z, An = (n−1, n+1); so A−1 = (−2, 0), A0 = (−1, 1),
A1 = (0, 2), etc. This does not contain a finite subcover.
In contrast, the set A1 = R is a finite open cover of R and contains a
finite subcover, itself.
(2) Consider the interval (0, 10). The open cover {An} of the real line given in the
previous example also covers this interval. This contains a finite subcover.
(3) Consider the interval (0, 10). The (countably infinite collection of) sets An =
(1/n, 10) is an open cover. This does not contain a finite subcover.
(4) Consider the interval (0, 10). The (uncountably infinite collection of) sets
defined for each x ∈ (0, 10), Ax = (x, 10) forms an open cover. This does not
contain a finite subcover.
(5) Consider the interval [0, 1]. Every open cover will contain a finite subcover.
Why?
12 3
Lemma 3.3.6. Suppose that A is a set with the property that every open cover of
A has a finite subcover. Then A is closed.
Suppose that A is not closed and a is a limit point of A that is not in A. Then
let An =
(−∞, a− 1
n
, a+ 1
n
,∞). We have that A ⊆ ⋃∞n=1An. However, because the
open sets in this cover are nested, if there was a finite subcover, it would follow that
there is some positive integer N so that A ⊆ AN . This would imply that if for all
x ∈ A, |x− a| ≥ 1
N
, and this contradicts that a is a limit point of A.
Lemma 3.3.7. Suppose that A is a set with the property that every open cover of
A has a finite subcover. If B ⊆ A is closed, then every open cover of B has a finite
subcover.
Suppose that B ⊆ ⋃α∈I Gα where each Gα is open is an arbitrary cover by open
sets. We have that A ⊆ B ∪Bc ⊆ (⋃α∈I Gα) ∪Bc is an open cover of A. This cover
has a finite subcover, and so there are finitely many sets G1, G2, . . ., Gn so that
A ⊆ G1 ∪G2 ∪ · · · ∪Gn ∪Bc.
It follows from this that
B ⊆ G1 ∪G2 ∪ · · · ∪Gn
and so the open cover of B has a finite subcover.
3. COMPACT SETS 13
Theorem 3.3.8 (Heine-Borel). A nonempty set K ∈ R is compact if and only if
every open cover of K has a finite subcover.
Suppose that K is a set with the property that every open cover has a finite
subcover. We proved above that K was closed. The set K must also be bounded
because K ⊆ ⋃∞n=−∞(n− 1, n+ 1) is an open cover. If K ⊆ (n1 − 1, n1 + 1) ∪ (n2 −
1, n2 + 1) ∪ · · · ∪ (nr − 1, nr + 1), then for all k ∈ K, |k| ≤ max{ni + 1 : 1 ≤ i ≤ r}
and so K is bounded.
Now suppose that K is closed and bounded. Then ∃M ∈ R so that K ⊆ [−M,M ].
Suppose that K ⊆ ⋃αGα is an open cover which does not have a finite subcover.
Then either K ∩ [−M, 0] or K ∩ [0,M ] must fail to have a finite subcover. Let
I1 = [−M, 0] or [0,M ] so that K ∩ I1 fails to have a finite subcover. If In−1 = [a, b] is
such that K∩In−1 fails to have a finite subcover, then either K∩[a, a+b2 ] or K∩[a+b2 , b]
fails to have a finite subcover, and define In so that Kn = K ∩ In fails to have a finite
subcover. This implies that Kn is nonempty. Thus, ∃x ∈ K ∩ (
⋂∞
n=1Kn). Since x is
in the open cover, there is some α so that x ∈ Gα and so there is some > 0 so that
B(x) ⊆ Gα. If n is large enough that the length of In < /2, then In ⊆ B(x) ⊆ Gα,
and this implies that K ∩ In does have a finite subcover, namely K ∩ In ⊆ Gα. This
contradicts the assumption that the open cover of K does not have a finite subcover.
Corollary 3.3.9. Any intersection (finite or infinite) of compact sets is compact.
If {Kα : α ∈ I} is a set of compact sets then
⋂
α∈I Kα is closed. Moreover, each
Kα is bounded, and a subset of a bounded set is also bounded.