# 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!

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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)