What would the formal proof look like for this problem?

Prove that the interior of an intersection is the intersection of the interiors:

(A∩B)º =Aº∩Bº

Printable View

- Feb 13th 2011, 01:45 PMalice8675309Interior points
What would the formal proof look like for this problem?

Prove that the interior of an intersection is the intersection of the interiors:

(A∩B)º =Aº∩Bº - Feb 13th 2011, 01:51 PMPlato
You have enough other postings here to know that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.

- Feb 13th 2011, 02:14 PMalice8675309
Sorry. However, this is not actually homework. I'm just trying to get comfortable with some topics i'm uneasy about. However, my lack of work posted is due to the fact that i seem to be confusing in my posts. however this is the idea i've been working with i just can't seem to put it into a formal proof. And I also don't even know if im going in the right direction. But i have this:

If we have an accumulation point p of A then there is a neighborhood of B that lies entirely inside of A and vise versa. I'm not sure how to put that into a proof and/or if my scratch work is even getting me anywhere. - Feb 13th 2011, 02:39 PMPlato
This question is about interior points not accumulation points.

The statement that $\displaystyle p$ is a interior point of a set $\displaystyle A$ means that there is a open set $\displaystyle \mathcal{O}$ such that $\displaystyle p\in\mathcal{O}\subset A$.

The symbol $\displaystyle A^o$ stands for the set of all interior points of $\displaystyle A$.

You want to show that $\displaystyle (A\cap B)^o=A^o\cap B^o.$

Interior points on LHS are interior points on RHS, and visa versa. - Feb 13th 2011, 04:30 PMDrSteve
Just a hint to get you started writing a formal proof:

A standard way to show that two sets are equal is to show that each is a subset of the other. So for example, start with an element of the left hand side, and argue that this element is also a member of the right hand side. In this problem the argument follows almost immediately from the definition of interior point (use Plato's post above to help with the details). - Feb 13th 2011, 04:50 PMalice8675309
Thanks! so I must show that p is an interior point of (A∩B) before I even look at the intersection part of the question right? So I would first write the definition of interior point, namely, that is a interior point of a set means that there is a open set O such that p$\displaystyle \in$O contained in A. And then from there I show that there is an x in (A∩B). So if x is in (A∩B) then x is in A and x is in B which means that x is in A ∩ B. And thus if p is an interior point of (A∩B) then it has to be an interior point of A∩B?

- Feb 13th 2011, 05:06 PMDrSteve
Not quite. Start like this:

Let $\displaystyle p$ be in the interior of $\displaystyle A\cap B$. By definition there is an open set $\displaystyle O\subseteq A\cap B$ with $\displaystyle p\in O$.

Can you take it from there? - Feb 13th 2011, 05:08 PMalice8675309
oh, i thought I also had to show that x was in the intersection. would it follow the same way?

- Feb 13th 2011, 05:23 PMDrSteve
You need to show that p is in the intersection of the two interiors. Can you finish the argument?

- Feb 13th 2011, 05:32 PMalice8675309
so it would be like let p be the interior point of (A∩B) then p is an interior point of A and and p is an interior point of B so hence p is an interior point of A∩B? (i know this is only the left hand side) but i guess i just felt like there was alot more going on. I'm sorry i'm really attempting to get my facts straight, hence why i always post so much, but i appreciate your help so much!

- Feb 13th 2011, 06:18 PMDrSteve
No. You have to start with an element of the set on the left and show it's an element of the set on the right:

Let $\displaystyle p\in (A\cap B)^o$. By definition there is an open set $\displaystyle O\subseteq A\cap B$ with $\displaystyle p\in O$. Since $\displaystyle A\cap B\subseteq A$, we have $\displaystyle O\subset A$. Similarly $\displaystyle O\subseteq B$. Thus $\displaystyle p$ is an interior point of $\displaystyle A$ and $\displaystyle p$ is an interior point of $\displaystyle B$. So $\displaystyle p\in A^o\cap B^o$.

Now start with an element in the right set and show it's in the set on the left. - Feb 13th 2011, 06:30 PMalice8675309
Thanks so much! You've been such a help!

- Feb 14th 2011, 08:51 AMalice8675309
For the other side of this proof ( proving from right to left) would be something like:

Let p$\displaystyle \in$(A* $\displaystyle \cap$B*) the p$\displaystyle \in$A* and p$\displaystyle \in$B* . Then p is in A and p is in B then p is in (A$\displaystyle \cap$B) which shows that A*$\displaystyle \cap$B* is a subset ? I'm not sure where to go from here or if what I'm saying makes any sense.By the way I had to use * symbol to indicate interior. - Feb 14th 2011, 09:16 AMDrSteve
You start off ok by applying the definition of the intersection. Once you know that p is in A* and B*, you now have to say what each of these means.

It is true that p is in each of A and B, but this is not helpful, because the interior of the intersection may be smaller than the intersection itself. - Feb 14th 2011, 09:31 AMalice8675309