Honors Introduction to Analysis I
Homework VI
Solution
March 23, 2009
Problem 1
Let
f
be a function defined on a closed domain
D
. Show that
f
is continous if and only if the inverse
image of every closed set is a closed set.
Solution.
“=
⇒
” Assume
f
is continuous. Let
A
be a closed set in the codomain. Then
A
c
is open, which means
that
f

1
(
A
c
) is also open (since
f
is continuous), so (
f

1
(
A
c
))
c
is closed.
But
f

1
(
A
) =
D
∩
(
f

1
(
A
c
))
c
, an
intersection of closed sets, so it’s closed.
“
⇐
=” Assume that if
A
is closed, then
f

1
(
A
) is closed.
For any
x
0
∈
D
and any
n
∈
N
, consider the set
A
= (
∞
, f
(
x
0
)

1
/n
]
∪
[
f
(
x
0
) + 1
/n,
∞
) = (
f
(
x
0
)

1
/n, f
(
x
0
) + 1
/n
)
c
, which is closed. Hence
f

1
(
A
) is also
closed.
By the construction
f
(
x
0
)
/
∈
A
, so
x
0
/
∈
f

1
(
A
).
Since
f

1
(
A
) is closed,
x
0
is not a limit point, so
there is a neighborhood of
x
0
not contained in
f

1
(
A
), i.e. there is some interval (
x
0

1
/m, x
0
+ 1
/m
) such that
(
x
0

1
/m, x
0
+ 1
/m
)
∩
f

1
(
A
) =
∅
. Then
f
((
x
0

1
/m, x
0
+ 1
/m
))
⊂
(
f
(
x
0
)

1
/n, f
(
x
0
) + 1
/n
) =
A
c
.
We have that
∀
n
, there is an
m
such that if

x

x
0

<
1
/n
, then

f
(
x
)

f
(
x
0
)

<
1
/m
, hence
f
is continuous.
Problem 2
Let
A
be the set defined by the inequalities
f
1
(
x
)
≥
0
,
f
2
(
x
)
≥
0
,...,
f
n
(
x
)
≥
0
, where
f
1
, f
2
, ..., f
n
are
continuous functions defined on the whole line.
Show that
A
is closed.
Show that the set defined by
f
1
(
x
)
>
0
,
f
2
(
x
)
>
0
,...,
f
n
(
x
)
>
0
is open.
Solution.
A
=
{
x
∈
R

f
k
(
x
)
≥
0
,
∀
k
= 1
, ..., n
}
=
n
\
k
=1
{
x
∈
R

f
k
(
x
)
≥
0
}
=
n
\
k
=1
f

1
k
([0
,
∞
))
Since [0
,
∞
) =
R
\
(
∞
,
0) is closed, by the previous problem,
f

1
k
([0
,
∞
)) is closed, for each
k
. So
A