# Point set proof

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## Homework Statement

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.

## Homework 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.

## 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.

jbunniii
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Gold Member
Do I assume the existence of a point in A0$\cup$B0 and then prove it is also contained within (A$\cup$B)0 ?
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 $x \in (A \cup B)^o$?

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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 $x \in (A \cup B)^o$?
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?

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Dick
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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?
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?

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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?
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?

Dick
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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?
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?

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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?
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?

Dick
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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?
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.

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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.
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?

Dick
Homework Helper
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?
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.

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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.
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}

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

jbunniii
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Gold Member
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.
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...

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

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Homework Helper
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...
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.

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

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

Dick
Homework Helper
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!

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

Dick
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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.

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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.

Dick
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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.

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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 ).

jbunniii
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Gold Member
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.