is this a good proof ? ( Real Analysis )

[wispy hemlock]

there was a theorem in the book and it proof.
i didn't see how the book proved it so i made an exercise for myself to prove it in my own way
so... is this a good proof ?

[rocky falcon]

A set S can only be closed, opened or both
Not true, a set doesn't have to be either.
Case 1... S = (a, b)
This set isn't closed, so it is not a "case" to look at for the theorem.
Case 2... S = [a, b)
This set is also not closed.

[wispy hemlock]

This set isn't closed, so it is not a "case" to look at for the theorem.
then how tf i would prove by contradiction ???
i need to show that all the other cases leads to a contradiction to prove that they're false

Not true, a set doesn't have to be either.
@rocky falcon so what is the name of the sets that takes the form [a,b) ?

Case 2 actually exists, what i made in the proof is making a wrong name, lmao

[rocky falcon]

They're calling this interval half-closed, but it isn't actually closed. It doesn't include the limit point 'a'.

For this proof you shouldn't need to use cases. What you do need is the definition of a closed set. What is the definition of closed in your book?

[wispy hemlock]

the set that contains all their limit points

[rocky falcon]

Ok, so in the foward direction (=>), you start by assuming S is closed, so that it contains all of its limit points. Then S^c doesn't contain any limit point of S. Why? Because if S^c contains a limit point of S, say x, then x ∈ S^c (by assumption) and x ∈ S (since x is a limit point of S and S is closed). But this is a contradiction. Therefore S^c contains no limit points of S if S is closed.