Set A={f element C([0,1]): f(x)>0 for all x in [0,1]}
Find the interior and closure of A in C([0,1])
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Set A={f element C([0,1]): f(x)>0 for all x in [0,1]}
Find the interior and closure of A in C([0,1])
Explain
Please help
i am looking at the definitions for interior and kind of confused because it says that in order for S to be interior, Int(S) is an open set of the metric space -X
looking at my metric space, C([0,1]) and my S, A, ...... it looks as thought that can't be true because C is a closed set
grrrrrr
A={f element C([0,1])Quote:
Originally Posted by derek walcott
when they write elements C([0,1]) that's means the elements of A is the continuous functions defined at [0,1]
so is.... C([0,1]) a line segment on a closed interval then?
and A is any point in C?
i can't visualize it
Int(A) = C((0,1))
and closure of A = C([0,1])
close... correct .... not close at all .....?
what is the whole question I think there is something missing,
what is the base for the space we have.
I solved it but I'm not sure ...
if we say A is the subset of the space that contains all continuous functions at [0,1]
and the functions of A is continuous and has a positive range
A is open in that space since all elements of A is continuous functions at [0,1]
so Int(A) =A
Closure of A = A if A is closed in the space
A is closed that means the complement of A is open
the complement of A is the continuous functions defined at [0,1]
with range equal 0 and negative range
as you can see the complement is open since it contains continuous functions defined at [0,1]
so Closure(A) = A since A is closed to
...
professor never said anything about the base of the space
ok what is the metric defined at the space ??
Let’s get the notation straight. Here is the usual understanding for this material.
is the set of continuous functions on
.
The metric is forthen
.
P.S.
Ifthe because of the properties of continuous functions we know that
has a positive minimum on
.
Now define.
Suppose
Can you show that