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.
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.
This question is about interior points not accumulation points.
The statement that is a interior point of a set means that there is a open set such that .
The symbol stands for the set of all interior points of .
You want to show that
Interior points on LHS are interior points on RHS, and visa versa.
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).
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 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?
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!
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 . By definition there is an open set with . Since , we have . Similarly . Thus is an interior point of and is an interior point of . So .
Now start with an element in the right set and show it's in the set on the left.
For the other side of this proof ( proving from right to left) would be something like:
Let p (A* B*) the p A* and p B* . Then p is in A and p is in B then p is in (A B) which shows that A* 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.
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.