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## Point set proof

 Quote by Zondrina Well A = {x$\in$ℝ | 0 ≤ x ≤ 5} B = {x$\in$ℝ | 5 ≤ x ≤ 10} So A0 = {x$\in$ℝ | 0 < x < 5} B0 = {x$\in$ℝ | 5 < x < 10} So we have : A0UB0 = {x$\in$ℝ | 0 < x < 10, x≠5} And also : (AUB)0 = {x$\in$ℝ | 0 < x < 10} Thus : A0UB0$\subset$(AUB)0
That's exactly what I wanted to hear. Thanks!

 Quote by Dick Works. Why does it work? Spell out the reason.
 Quote by Dick That's exactly what I wanted to hear. Thanks!
Oh man thanks so much for your patience, really though. I just wanted to understand this so badly.

Also my final concern, is what jbuni said true? Am I too close? Or was my proof sufficient?

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 Quote by Zondrina Oh man thanks so much for your patience, really though. I just wanted to understand this so badly. Also my final concern, is what jbuni said true? Am I too close? Or was my proof sufficient?
What you said convinces me you understand it. jbunnii's rephrasing is a better version for the proof.

 Quote by Dick What you said convinces me you understand it. jbunnii's rephrasing is a better version for the proof.
 You have a neighborhood N of x such that x⊂N⊂A. Furthermore, A⊂A∪B, so it follows that x⊂N⊂A∪B. Therefore..
Therefore since A is contained within AUB, it follows that A0 is contained within (AUB)0.

I believe that's what he meant to say. If so then I do agree it's a better way to phrase this.

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 Quote by Zondrina Therefore since A is contained within AUB, it follows that A0 is contained within (AUB)0. I believe that's what he meant to say. If so then I do agree it's a better way to phrase this.
Yes, if the neighborhood N of x is in A, then it's certainly in AUB.

 Quote by Dick Yes, if the neighborhood N of x is in A, then it's certainly in AUB.

Perfect, thanks again for all your help man. Though I wish my professor didn't dive right into topology as soon as the class started... ( Its only calc II lol ).

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 Quote by Zondrina Therefore since A is contained within AUB, it follows that A0 is contained within (AUB)0. I believe that's what he meant to say. If so then I do agree it's a better way to phrase this.
Just to follow up on my earlier post...

You have a neighborhood N of x such that $x \subset N \subset A$. Furthermore, $A \subset A \cup B$, so it follows that $x \subset N \subset A \cup B$. Therefore x is an interior point of $A \cup B$, i.e. $x \in (A \cup B)^o$. Since $x$ was an arbitrary point of $A^o$, this shows that $A^o \subset (A \cup B)^o$.

(Then argue similarly for $x \in B^o$ and wrap up the proof.)

That's the level of detail I would like to see if I were grading this problem, so I could be confident that you understood why every step was true.

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