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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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
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
...
Let’s get the notation straight. Here is the usual understanding for this material.
$\displaystyle \mathcal{C}([0,1])$ is the set of continuous functions on $\displaystyle [0,1]$.
The metric is for $\displaystyle \{f,g\}\subset\mathcal{C}([0,1])$ then $\displaystyle d(f,g) = {\max }_{x \in [0,1]}\left| {f(x) - g(x)} \right|$.
P.S.
If $\displaystyle f\in \mathcal{C}([0,1])\cap A$ the because of the properties of continuous functions we know that
$\displaystyle f$ has a positive minimum on $\displaystyle [0,1]$. Lets call it $\displaystyle f(t)$.
Now define $\displaystyle \varepsilon = \frac{{f(t)}}{2} > 0$.
Suppose $\displaystyle h\in\mathcal{B}(f; \varepsilon )=\{g:d(f,g)< \varepsilon \}$
Can you show that $\displaystyle h\in A?$