# interior/ closure

• April 25th 2010, 12:10 PM
derek walcott
interior/ closure
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

• April 25th 2010, 12:24 PM
derek walcott
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
• April 25th 2010, 12:29 PM
Amer
Quote:

Originally Posted by derek walcott
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])

when they write elements C([0,1]) that's means the elements of A is the continuous functions defined at [0,1]
• April 25th 2010, 12:55 PM
derek walcott
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
• April 25th 2010, 01:23 PM
derek walcott
Int(A) = C((0,1))
and closure of A = C([0,1])

close... correct .... not close at all .....?
• April 25th 2010, 01:27 PM
Amer
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

...
• April 25th 2010, 01:33 PM
derek walcott
professor never said anything about the base of the space
• April 25th 2010, 01:35 PM
Amer
ok what is the metric defined at the space ??
• April 25th 2010, 01:54 PM
Plato
Let’s get the notation straight. Here is the usual understanding for this material.
$\mathcal{C}([0,1])$ is the set of continuous functions on $[0,1]$.
The metric is for $\{f,g\}\subset\mathcal{C}([0,1])$ then $d(f,g) = {\max }_{x \in [0,1]}\left| {f(x) - g(x)} \right|$.

P.S.
If $f\in \mathcal{C}([0,1])\cap A$ the because of the properties of continuous functions we know that
$f$ has a positive minimum on $[0,1]$. Lets call it $f(t)$.
Now define $\varepsilon = \frac{{f(t)}}{2} > 0$.
Suppose $h\in\mathcal{B}(f; \varepsilon )=\{g:d(f,g)< \varepsilon \}$
Can you show that $h\in A?$