# Topology (Boundary points, Interior Points, Closure, etc...)

Tags: boundary, closure, interior, points, topology
 P: 188 Hi. Can somebody please check my work!? I'm just not sure about 2 things, and if they are wrong, all my work is wrong. 1. Find a counter example for "If S is closed, then cl (int S) = S I chose S = {2}. Im not sure if S = {2} is an closed set? I think it is becasue S ={2} does not have an interior point, and 2 has to be a boundary point, (and 2 cannot be both and interior and a boundary point.) Im sure S={2} is closed! cl (int S) =cl (int 2) =cl (empty) = empty and that is not equal to S = {2} 2. Let A be a nonempty open subset of R and let Q be the set of rationals. Prove that (A n Q)   .... (I hope those symbols show, I got them from MS Word) I figured that Since "A is a nonempty open subset of R," A has to be composed of MORE than 1 element, hence, A has to have 2,3,4.... elements. And I know (from lectures and the text book) that between any 2 real numbers, their is a rational. Hence, (A n Q)  . How does that sound? Can somebody please check this? Thanks in advance.
 HW Helper P: 2,566 Your answer to 1 sounds good, as long as your considering {2} as a subset of R with the usual topology. For 2, I assume by your answer that you want to show: $$A \cap Q \neq \emptyset$$ To be more rigorous, you should show that A must contain an open interval and that any open interval in R must contain some rational numbers.
P: 188
 Quote by StatusX Your answer to 1 sounds good, as long as your considering {2} as a subset of R with the usual topology. For 2, I assume by your answer that you want to show: $$A \cap Q \neq \emptyset$$ To be more rigorous, you should show that A must contain an open interval and that any open interval in R must contain some rational numbers.

Thank you for confirming my answers!

Yes, for 2), I was trying to find $$A \cap Q \neq \emptyset$$ ... but im sure Im on the right track for that!

HW Helper
P: 9,395

## Topology (Boundary points, Interior Points, Closure, etc...)

Defintion of open:

A is open if for any point a in A there is an interval (x-e,x+e) contained in A for some e>0 (e depends on a).

Now, you need to show that this has a rational number in it, which can be done in many ways of varying highbrow-ness.
P: 188
 Quote by matt grime Defintion of open: A is open if for any point a in A there is an interval (x-e,x+e) contained in A for some e>0 (e depends on a). Now, you need to show that this has a rational number in it, which can be done in many ways of varying highbrow-ness.
Thats right! i never thought of it like that.

A is open if all the points in A are interior points, and a point x in A is an interior point if a Neighbourhood of x is contained in A ... just like you said, (x-e, x+e).

Since A is a subset of the Real Line, it contains Q.

I have another Question. Prove: An accumulation point of a set S is either an interior point of S or a bounadry point of S

Would it be okay if I write: "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point."

I personally find my above "If - Then" statement MUCH easier to prove!
HW Helper
P: 9,395
 Quote by rad0786 Thats right! i never thought of it like that. A is open if all the points in A are interior points, and a point x in A is an interior point if a Neighbourhood of x is contained in A ... just like you said, (x-e, x+e). Since A is a subset of the Real Line, it contains Q.

that is a huge leap. It contains an element of Q (actually it contains infintely many elements of Q), it doesn't contain Q.

 I have another Question. Prove: An accumulation point of a set S is either an interior point of S or a bounadry point of S Would it be okay if I write: "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point." I personally find my above "If - Then" statement MUCH easier to prove!
It is an 'if then' statement already, however it is exactly the reverse of what you wrote.
P: 188
 Quote by matt grime that is a huge leap. It contains an element of Q (actually it contains infintely many elements of Q), it doesn't contain Q.
Oops, that was a hudge leap. But I got the idea that it contains an element of Q.
I could say that between any two real numbers in the interval A their is a rational number (As prooved in lecture), thus, their exsits a rational number in A

 Quote by matt grime It is an 'if then' statement already, however it is exactly the reverse of what you wrote.
So if I prove what I worte, then its as if I assume its an "if and only if statement" which would be an invalid assumption.
 HW Helper Sci Advisor P: 9,395 Erm, you are free to prove the statement you made, however it doesn't at all answer the question you were asked. Rather than say 'it is as if I assume it is an if and only if statement', I would say you have just not proved what you were asked to prove. I don't know what you were assuming.
 P: 188 You are right. Oh, what I was trying to say was, for example, If it rains, then I'll watch TV If I watch TV, then it rains. These 2 statements don't mean the same thing. But If I want to prove "it rains if and only if I watch TV" Then ill have to prove If it rains, then I'll watch TV If I watch TV, then it rains. So for the question, I was trying to prove it the wrong way beacue it was easier, however, I can't do that. So ill just have to prove "An accumulation point of a set S is either an interior point of S or a bounadry point of S" and NOT "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point." :)
HW Helper
P: 1,996
 Quote by rad0786 So ill just have to prove "An accumulation point of a set S is either an interior point of S or a bounadry point of S" and NOT "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point." :)
How did you manage to prove your version? It looks false to me.
 P: 188 Oh no.. i think im confusing you guys.. im Sorry for that. I didnt prove it yet, I was just wondering if it would be valid to proove "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point." INSTEAD of proving "An accumulation point of a set S is either an interior point of S or a bounadry point of S" ... Which i found out you can't do that Sorry for the confusion.
HW Helper
 P: 188 Okay, I've spent all of yesterday and most of today on this same question. With very little progress. "An accumulation point of a set S is either an interior point of S or a bounadry point of S" Suppose $a \in S'$ (S' is the set of accumulation points). Then $a \in (N^* (x,e) \cap S) \neq \emptyset$ ... which means that $a \in S$. Since $a \in S$, then a is either an interior point or a boundary point? How does that sound? I feel like its not good.. but thats as far as I could get. I tried breaking it up into cases, case1: S is open, case2: S is closed.... where case1 would mean its an interior point. But this fails because i cannot come up with anything for case2 Anybody got any ideas please?