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Proof of a relationship between interior and closure

  • Thread starter jaci55555
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  • #1
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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
 

Answers and Replies

  • #2
360
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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 [itex] x \in cl(A) [/itex]. Now we need to show that [itex] x \notin (X\backslash A)^\circ[/itex]. For contradiction, suppose that [itex] x \in (X\backslash A)^\circ [/itex]. Then there is some [itex]r_0 >0[/itex] such that [itex] B(x, r_0) \subseteq X \backslash A [/itex]. Thus [itex] B(x, r_0) \cap A = \varnothing[/itex]. However, this contradicts the assumption that [itex] x \in cl(A) [/itex] since for any r>0, we have [itex] B(x, r) \cap A \neq \varnothing[/itex] by the definition of [itex] cl(A) [/itex]. Therefore [itex] cl(A) \subseteq X \backslash (X \backslash A)^\circ[/itex].

Did that make sense?

If so, then try the other implication! :smile:
 
Last edited:
  • #3
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Oh and welcome to the forum!
 
  • #4
29
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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)
 
  • #5
360
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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.
 
  • #6
29
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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)
 

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