1. The document proves that for a closed and bounded above subset F of the real numbers, the supremum of F exists and is equal to the maximum of F. It does this by showing that the subsets {x in F: x >= h} are compact for h < supF, and thus their intersection is non-empty. 2. It proves that the set {0,1,1/2,1/3,...} is compact as any sequence in it has a convergent subsequence, but the set {1,1/2,1/3,...} is not compact as the sequence {1/n} has no convergent subsequence within the set. 3. It

1. #1. Let F be a closed bounded above subset of ¡ . Prove that max F exists.
Since F is bounded above, there exists supM F= . By definition of supremum, for any
h<M the subset
(1) { }:hF x F x h= ∈ ≥
is not empty.
We’ll prove that it is a compact subset. Since hF F⊆ and F is closed set, we have
{ }: [ , ]hF x F h x M F h M= ∈ ≤ ≤ = I so that hF is an intersection of closed set F and a
compact interval [h, M] and, therefore, hF , h<M , is a compact subset in ¡ .
From definition (1) we also have
(2) M>h k> ⇒ h kF F⊆ .
Let’s define a sequence
1
nh M
n
= − , n=1,2,3,…, so that nh M< and subsets hn
F are
closed. From (2) we assert that the sequence { } 1
hn
n
F
∞
=
is sequence of compact nested
subsets: m>n ⇒ m nh h> (using 2) ⇒ h hm n
F F⊆ . By theorem of nested compact sets,
there exists a point that belongs to the intersection,
1
hn
n
p F
∞
=
∈I . This means that
, 1,2,...hn
p F F n∈ ⊆ = , and the inequality
1
holds for n=1,2,...M p M
n
− ≤ ≤
.
We assert that p M F= ∈ , which proves that M=supF belongs to F, and therefore,
M = sup F = max F .
#2. Prove that set {0,1,1/2,1/3,… } compact but the set {1,1/2,1/3,… } is not.
1. The set X={0,1,1/2,1/3,, } is compact because any sequence in it has convergent
subsequence.
Proof.
Let { } 1i i
x
∞
=
be some sequence in the set X={0,1,1/2,1/3,… }.
Case1. Let’s assume that there exists an element x that has infinite count within in the
sequence { } 1i i
x
∞
=
. That means that there exists an infinite subsequence of indices ni ,
n=1,2,…, such that in
x x= . Since the sequence { } 1
in n
x
∞
=
is constant, it converges, its limit
is equal to x and belongs to X.
1

2. Case 2.
Let’s assume that any element x has finite count within in the sequence { } 1i i
x
∞
=
. In this
case
Subset of different elements in the sequence is infinite and for any n>0
there exists element in
x of the sequence such that
1
, 1,2,3,...in
x n
n
≤ = .
Since
1
0 in
x
n
≤ ≤ , we assert (sqeezing theorem of limits) that limit of subsequence
{ } 1
in n
x
∞
=
exists and lim 0inn
x
→∞
= X∈ .
2. The set Y={1,1/2,1/3,… } is not compact because there exists a sequence any
subsequence of which does not converge in Y.
Let’s consider sequence
1
1
nn
∞
=
 
 
 
. Its has limit 0 which does not belong to the set Y.
Since any subsequence of this sequence has the same limit we conclude that any
subsequence of
1
1
nn
∞
=
 
 
 
does not converge in Y.
#3
Let f be a continuous function →¡ ¡ , and K be a compact subset of ¡ .
Prove that ( )f K is a compact set.
Proof.
Let i
i
CU is an arbitrary cover of ( )f K with open sets { }iC . For any iC let
{ }1
[ ] : ( )i if C x f x C−
= ∈ ∈¡ be an inverse image of the set iC . Since the function f is continuous, sets
1
[ ]if C−
are open in ¡ . Also,
1 1 1
[ ] [ ] [ ( )]i i
i i
f C f C f f K K− − −
= ⊇ ⊇U U , so that
sets
1
[ ]if C−
make an open cover of the set K. Since K is a compact set in ¡ , there exists
finite subcover
1
1
[ ]
N
in
n
f C K−
=
⊇U . Using
1
( [ ])i in n
f f C C−
⊆ we assert
1 1
1 1 1
( ) [ ] ( [ ])
N N N
i i in n n
n n n
f K f f C f f C C− −
= = =
 
⊆ = ⊆ ÷
 
U U U .
This means that an arbitrary cover of ( )f K with open sets { }iC has finite subcover and, therefore, the
set ( )f K is a compact set.
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