If i 2 s is finite switch the names of i 2 and j 2

a contradiction). If I
2
∩ S is finite, switch the names of I
2
and J
2
. Now I
2
∩ S has
infinitely many points. Divide I
2
into two equal intervals I
3
and J
3
. Continue in
this way, producing a sequence of nested closed intervals I
1
⊇ I
2
⊇ I
3
. . . such that
I
n
∩ S has infinitely many elements ∀ n ∈ N . Write I
n
= [ a
n
, b
n
].
By the nested intervals theorem,
∞
intersectiontext
n =1
I
n
negationslash = ∅ . Let x ∈
∞
intersectiontext
n =1
I
n
. Claim: x is an
accumulation point of S . To see this, given any ǫ > 0 , ∃ n ∈ N s.t.
1
n
<
ǫ
2 K
(using
(c) of the Archimedian property). Now n ≤ 2
n
by mathematical induction, so
1
2 n
≤
1
n
<
ǫ
2 K
. Hence
2 K
2 n
< ǫ . Since the length of [ a
n
, b
n
] is
2 K
2 n
, we have
b
n
= a
n
+
2 K
2 n
< a
n
+ ǫ ≤ x + ǫ,
(recall that x ∈ [ a
n
, b
n
]). Similarly,
x − ǫ ≤ b
n
− ǫ < b
n
−
2 K
2 n
= a
n
.
So [ a
n
, b
n
] ⊂ ( x − ǫ, x + ǫ ). Since [ a
n
, b
n
] contains infinitely many points in S , so
does ( x − ǫ, x + ǫ ). Since ǫ > 0 was arbitrary, x is an accumulation point of S by
Lemma 3.22. square
Homework 3/4. (Go by the webpage for the precise numbers and due dates)
(1) Show that the interior of a set is open. Also show that Int( S ) is the largest
open set contained in S .
(2) Show that
¯
S is the smallest closed set containing S .
(3) Homework 3.5 Exercise 1 from the textbook 5th Ed (Exercise 14.1 in 4th
Edition).
(4) Explain why each of the following sets are not compact: (a) [1 , 3), (b) N ,
(c) { 1 /n : n ∈ N } , (d) { x ∈ Q : 0 ≤ x ≤ 2 } .
(5) Which of the 9 types of intervals are compact?
(6) Show that any isolated point of S is in Bdy(S).
(7) Show that the closure of a bounded set is bounded.
(8) Read the proofs above in this Section which we skipped in class.
END OF WORK FOR TEST 2
Appendix to Chapter 3. Proof of the Heine-Borel theorem.
THE PROOF BELOW IS NOT EXAMINABLE
We define an open cover of a set S to be a collection C of open sets whose union
contains S . A finite subcover of S from C is a finite collection { C
1
, C
2
, · · · , C
n
} of

MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES 31
the sets in C , whose union contains S . That is, C
k
∈ C for all k , and S ⊂ ∪
n
k =1
C
k
.
Note that if C
1
⊂ C
2
⊂ · · · ⊂ C
n
then C
n
= ∪ n
k =1
C
k
, so S ⊂ C
n
.
Example. Let S = [0 , 1] and let
C = { ( −
1
50
,
1
50
) , (
1
2
,
3
2
) , (
1
4
,
3
4
) , (
1
8
,
3
8
) , (
1
16
,
3
16
) , (
1
32
,
3
32
) , (
1
64
,
3
64
) , (
1
128
,
3
128
) , · · · } .
It is easy to see that in this last example, C is an infinite collection of open sets
whose union contains S . That is, C is an open cover of S . But in fact the first 7
sets listed in C have a union which already contains S . These 7 sets comprise a
finite subcover. This example also illustrates the condition in (2) of the Heine-Borel
Theorem.
Theorem 3.30. (The Heine-Borel theorem) If S is a nonempty set of real numbers
then the following are equivalent:
(1) S is closed and bounded.
(2) S has the following property: whenever C is a collection of open sets whose
union contains S , then actually S is contained in the union of a finite
number of the sets in C . (In English: every open cover of S has a finite
subcover.)
Proof. (2) ⇒ (1) Let C = { ( − n, n ) : n ∈ N } . By (2), this collection has a finite