Proof of a relationship between interior and closure

[jaci55555]

A^closure = X\(X\A)^interior

I am REALLY bad at proofs. I never know where to start. I only have the definitions of closure and interior. I feel like they threw us in the deep end

I've written like 3pages, but mostly just pictures.
interior: a is an element of A^int iff there exists r>0 with B(a,r) subset of A
and
closure: a is an element of A^closure iff for all r>0, B(a,r)(and)A are not empty

 

[spamiam]

I'll do one implication, you try the other.

(Edit: \overline isn't working in TeX, so I'll denote the closure of A by cl(A).)

Suppose x \in cl(A). Now we need to show that x \notin (X\backslash A)^\circ. For contradiction, suppose that x \in (X\backslash A)^\circ. Then there is some r_0 >0 such that B(x, r_0) \subseteq X \backslash A. Thus B(x, r_0) \cap A = \varnothing. However, this contradicts the assumption that x \in cl(A) since for any r>0, we have B(x, r) \cap A \neq \varnothing by the definition of cl(A). Therefore cl(A) \subseteq X \backslash (X \backslash A)^\circ.

Did that make sense?

If so, then try the other implication!

 

[spamiam]

Oh and welcome to the forum!

 

[jaci55555]

Thank you :shy:
Yours totally makes sense! How did you know to do it using a contradiction?
I tried a similar thing, I hope it's right:

Suppose x element of X\int(X\A), thus x not an element of int(X\A).
We need to show that x element of closure(A).
By contradiction, suppose that x not element of clos(A). Then there is some r>0 such that B(x, r)(and)A=empty set. So that B(x, r) a subset of X\A.
However, this contradicts the assumption since x not an element of int(X\A)
Therefore x an element of clos(A)

 

[spamiam]

It's often useful to do a proof by contradiction when you're trying to prove a negative statement, and in this case a was trying to show that x was not an element of a set. When trying to prove a negative statement, you typically either have to rephrase it as a positive statement, or use contradiction. I'll just say that is considered by some to be "inelegant" to use a proof by contradiction except when totally necessary.

Your proof looks correct to me! Since in this case you're actually proving a positive statement, you may want to prove it directly, i.e. without using contradiction. It should boil down to the same thing in the end, though.

 

[jaci55555]

Thank you so much! I tried some other ways, but I'm not sure about what I can conclude using only x is not an element of int(X-A)