# Point set proof

1. Sep 14, 2012

### Zondrina

1. The problem statement, all variables and given/known data

Let A and B be subsets of ℝn with A0, B0 denoting the sets of interior points for A and B respectively. Prove that A0$\cup$B0 is a subset of the interior of A$\cup$B. Give an example where the inclusion is strict.

2. Relevant equations

I know a point Q$\in$S is an interior point of S if $\exists N_δ(Q)$ which is a subset of S.

3. The attempt at a solution

I've never actually attempted a problem like this, I'm wondering where to start really. Do I assume the existence of a point in A0$\cup$B0 and then prove it is also contained within (A$\cup$B)0 ? Any nudge in the right direction would be very helpful.

2. Sep 14, 2012

### jbunniii

Yes, that's exactly what you need to do.

So, suppose $x \in A^o \cup B^o$. Then either $x \in A^o$ or $x \in B^o$. If $x \in A^o$, can you prove that $x \in (A \cup B)^o$?

3. Sep 14, 2012

### Zondrina

EDIT : I thought about it a bit, is this better?

Suppose we pick x$\in$A0. WAIT NOW... We know that A0 is the set of interior points for A so since x$\in$A0 we can also say that x$\in$A since x is contained within the interior of A. If x$\in$A, then it follows that x$\in$(A$\cup$B) and will be contained within the interior of A$\cup$B that is x$\in$(A$\cup$B)0?

I know this is wrong, but maybe its a step in the right direction?

Last edited: Sep 14, 2012
4. Sep 14, 2012

### Dick

Use the definition of interior point. If x is an interior point of A then there is a neighborhood of x that is contained in A. Is that neighborhood contained in AUB?

5. Sep 14, 2012

### Zondrina

It seems trivial in saying that. Showing it is a different story though.

So since x$\in$A0, $\exists δ>0 | N_δ(x) \subseteq A$

This is obvious to me ^, but how would I use this notion to show that x$\in$(A$\cup$B) or does it go without saying?

6. Sep 14, 2012

### Dick

Goes through without much said. If x is in the interior of A then x is in A. x is in the neighborhood. If x is in A then x is in AUB. Now what about the neighborhood of x?

7. Sep 14, 2012

### Zondrina

Since x is contained within the interior of A, we know x is contained within A itself. The neighborhood of x is therefore contained within A.

Since x is in A, we know that x is in A$\cup$B

So would it follow that the neighborhood of x is contained ( is a subset of ) within A$\cup$B? If this were the case then since we know the neighborhood is inside A$\cup$B, we know it would be contained within the interior of A$\cup$B?

8. Sep 14, 2012

### Dick

You are making this seem harder than it is. If the neighborhood of x is contained in A (by definition of x being an interior point), then it must be contained in AUB, right? A is contained in AUB.

9. Sep 14, 2012

### Zondrina

My apologies, it's been a long day so my brain isn't registering like it usually does.

Yes i understand that the neighborhood of x being contained in A tells us that the neighborhood is also contained within A$\cup$B.

So if A $\subseteq$ A$\cup$B that tells me that A0 $\subseteq$ (A$\cup$B)0? Is it actually that straightforward?

10. Sep 14, 2012

### Dick

Yes, it is. Now you have to find the example where the inclusion is strict. You can find one by thinking about closed intervals on the real line.

11. Sep 14, 2012

### Zondrina

So ill just clean up everything that's been said into one post here :

Suppose x$\in$(A0$\cup$B0). Then x$\in$A0 or x$\in$B0.

Suppose x$\in$A0. Then x is an interior point of A and $\exists δ>0|N_δ(Q) \subseteq A$

Since the neighborhood of x is contained within A, it follows that it is also contained within A$\cup$B.

Now since A$\subseteq$A$\cup$B, it follows that A0$\subseteq$(A$\cup$B)0. If we had chosen x$\in$B0 we would have come to the conclusion that B0$\subseteq$(A$\cup$B)0 through the exact same logic. So it follows that A0B0$\subseteq$(A$\cup$B)0 as desired.

Now as for the example where the inclusion is strict would I need something like :

A = {x$\in$ℝ | 0 ≤ x ≤ 1} so A0 = {x$\in$ℝ | 0 < x < 1}

12. Sep 14, 2012

### Dick

Seems ok for the proof. To show the inclusion may be strict, you need to define B as well, yes?

13. Sep 14, 2012

### jbunniii

I would be careful with how you've worded this.

Indeed, "since A$\subseteq$A$\cup$B, it follows that A0$\subseteq$(A$\cup$B)0" is very close to the statement of what you have been asked to prove, without the proof.

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...

14. Sep 14, 2012

### Zondrina

Yes yes, just ensuring I needed something of that sort. Okay hmm, strict inclusion means there should be at least one element not included so would ...

A = {x$\in$ℝ | 0 ≤ x ≤ 5}
B = {x$\in$ℝ | 5 ≤ x ≤ 10}

Work?

Last edited: Sep 14, 2012
15. Sep 14, 2012

### Zondrina

Therefore the neighborhood of x is also contained within the interior of the union of A and B? Also I believe what I said is sufficiently far enough from the original question, but how would I rephrase this portion.

16. Sep 14, 2012

### Dick

Works. Why does it work? Spell out the reason.

17. Sep 14, 2012

### 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

18. Sep 14, 2012

### Dick

That's exactly what I wanted to hear. Thanks!

19. Sep 14, 2012

### 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?

20. Sep 14, 2012

### Dick

What you said convinces me you understand it. jbunnii's rephrasing is a better version for the proof.

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