18.100B/C: Fall 2008
Solutions to Homework 3
1.
By condition (a),
E
is nonempty. By condition (b),
E
is open. By condition (c), every point
that is a limit of a sequence of points in
E
is itself in
E
. Note that if
e
is a limit point of
E
, we
can choose a sequence of points of
E
whose limit is
e
: choose
x
n
to be a arbitrary point of
E
that
lies in the ball
B
(
e,
1
n
) and then lim
n
→∞
x
n
=
e
. Thus,
E
contains all its limit points and so
E
is closed. By Homework 2, Problem 8, a nonempty open and closed subset of a connected metric
space must be the entire space, so
E
=
X
as desired.
2.(a)
Equivalent: Suppose
x
n
→
p
. Then for all
r >
0 there is some
M
∈
N
such that
n
≥
M
implies
d
(
x
n
, p
)
< r
.
Choose any
N
and define
r
=
N

1
, and with the corresponding
M
set
ε
=
M

1
, so
M
=
ε

1
.
Then we have that for all
n
≥
ε

1
,
d
(
x
n
, p
)
< N

1
.
Thus, for every
N
there is some
ε >
0 such that for all
n
≥
ε

1
,
d
(
x
n
, p
)
< N

1
and so statement (a) is true.
Conversely, suppose that statement (a) holds of a sequence
x
n
and a point
p
. Then for all
M
∈
N
there is some
r >
0 such that for all
n
≥
r

1
,
d
(
x
n
, p
)
< M

1
. Choose any
ε >
0 and choose some
M
≥
ε

1
, so
M

1
≤
ε
. For this value of
M
these is an associated value of
r
, and pick some integer
N
≥
r

1
.
Then for all
n
≥
N
we have
d
(
x
n
, p
)
< M

1
≤
ε
.
Thus, for all
ε >
0 there is some
N
∈
N
such that
n
≥
N
implies
d
(
x
n
, p
)
< ε
and so
x
n
→
p
. (The moral of this question: variables
are just placeholders, and there’s nothing special about the letters
N
and
ε
.)
(b)
Not equivalent: consider
X
=
R
and
x
n
=
1
n
.
Then lim
n
→∞
x
n
= 0.
However, for any
N
∈
N
, with
ε
=
1
N
+1
and
n
=
N
we have
d
(
x
n
,
0) =
d
(
x
N
,
0) =
1
N
>
1
N
+1
=
ε
, so the given
statement is false. (In fact, statement (b) is equivalent to the statement “
x
n
is eventually constant.”)
(Moral: quantifiers matter.)
(c)
Not equivalent: consider
X
=
R
and the sequence
x
n
defined by
x
2
n
=
1
n
,
x
2
n

1
=
n
. (The
first few terms of this sequence are 1
,
1
,
2
,
1
2
,
3
,
1
3
, . . .
.) The range of this sequence has a single limit
point at 0, but the sequence
x
n
has no limit. (Another counterexample is the sequence defined by
x
2
n
=
1
n
,
x
2
n

1
= 1, since 1 is a limit of a subsequence, but an isolated point of the range.)
(d)
Not equivalent:
consider
X
=
R
and
x
n
= (

1)
n
, then the range
{
1
,
1
}
is finite and
hence
B
r
(
p
)
C
∩ {
1
,
1
}
is finite for all
r >
0. However, the sequence is known to diverge. (On the
other hand, the statement “For all
r >
0 the set
{
n
∈
N

x
n
∈
B
r
(
p
)
C
}
is finite.” is equivalent to
convergence; see Rudin 3.2(a).)
3.
Suppose
s
n
→
z
. Then for all
ε >
0 there is some
N
∈
N
such that for all
n
≥
N
we have
d
C
(
s
n
, z
)
< ε
. But
d
C
(
s
n
, z
) =

s
n

z
 ≥ 
s
n
  
z

=
d
R
(

s
n

,

z

), so for all
ε >
0 there is some
N
∈
N
such that for all
n
≥
N
we have
d
R
(

s
n

,

z

)
< ε
, i.e.,

s
n
 → 
z

and so

s
n

converges.