# M317 sec 2 Assignment 3 due Wed 1. Prove that if ```M317 sec 2
Assignment 3
due Wed
1. Prove that if
inf A exists, but
A, then must be an accumulation point for A.
Use this fact to prove that every finite set contains its inf.
Let
inf A
A. Then is a lower bound for A and for every
0,
is not a LB
p
. Since
A and p A, it
for A. Then there is some element p A such that
can’t be that
p, so
p
. But this implies that
0, N
A
so is an
accumulation point of A. Any finite set is composed entirely of isolated points so it has no
accumulation points. Therefore the inf must belong to the set. In fact it is just the smallest
element in the set.
2. Prove that if inf A exists then it is unique.
for any other
Let 1 inf A and 2 inf A . Then 1 is a lower bound for A and 1
lower bound, . In particular, 1
2 . By the same arguement, 2
1 . But then 1
2.
3. Prove that if A and B are closed, then A
B is closed.
A B is closed if A B C is open. We know that A B C A C B C and since A and B
are closed, it follows that A C and B C are open so that every point of A C and every point of B C
is an interior point. Let p denote any point of A C B C . Then p belongs to A C or B C , so
suppose p A C . Then p is an interior point of A C so it is also an interior point of A C B C .
This shows that every point of A C B C is an interior point so A C B C is open and A B is
closed.
m
r n , where 0
4. Let S m
n 0
above by the value
1
r
1. Show that the set of values A
1 .
r
We showed in class that S m
rS m
Sm
5. To show that
S 1 , S 2 , . . . is bounded
1 r m 1 , and this implies
1 rm 1
1 .
1 r
1 r
sup A , show that for every
0,
Sm
, if m
log
1 r
log r
.
Note that
0
Sm
rm 1
1 r
Then
if
For 0
r,
1 we have that 0
Sm
rm 1
1 r
rm 1
1
r ,
, if
1
m
log
1 r
log r
.
2
```