4.4: 8, 11, 12
Proof
. It is considered impossible. Because
f
is continuous and [0,1] is compact, by
Theorem 4.4.1, the range is compact as well. However, (0,1) is not considered compact.
Better said, there cannot exist a continuous function defined on [0,1] that possesses some
range (0,1). Since [0,1] is a compact set and continuous image of the compact srt [0,1]
will be compact while the range set (0,1) is not compact. QED.
Proof
. Let
f
: (0,1)
, which would be given by
f
(
x
) =
+
sin
with
→ [0, 1]
1
2
1
2
(2π?)
. So, the function
f
is continuous and onto. Therefore, the set (0,1) has some
? ∈ (0, 1)
range [0,1] under such a map as mentioned above. QED.
Proof
. There cannot be some continuous function from the set (0,1] with range set (0,1).
Now, if it is possible, we let
, which implies that
?((0, 1] )
= (0, 1)
?
−1
(0, 1)
=
(0, 1] .
This can be considered a contradiction, since
is open as
f
a continuous map, but
?
−1
(0, 1)
(0,1] is not open. QED.
Proof
. (
) Let us make the assumption that
is indeed continuous on
R
and let
R
be
⇒
?
𝑂 ⊆
open. We yearn to show that
is open. Then, we fix
and demonstrate
?