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

rad0786

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.

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:

[tex] A \cap Q \neq \emptyset[/tex]

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.

rad0786

Thank you for confirming my answers!

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

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.

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.


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!

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.


It is an 'if then' statement already, however it is exactly the reverse of what you wrote.

rad0786

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


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.

matt grime

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.

rad0786

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." :)

shmoe

How did you manage to prove your version? It looks false to me.

rad0786

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.

shmoe

I did understand this point. I'm just pointing out that the statement "If a point in S is either an interior point of S or a boundary point of S, then it is an acculumation point." is false, irregardless of the fact that it wouldn't solve your question. Your set S={2} will serve as a counter example, 2 is a boundary point yet not an accumulation point (at least not under the definition I'm used to, i.e. every neighbourood of 2 would need a point in S\{2}).

rad0786

Ohhh okay! I see what you'r saying. That is true indeed! I learned now that I have to be very carefull when doing proofs like this ... and their is not to look for an easy way out.

rad0786

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 [itex]a \in S'[/itex] (S' is the set of accumulation points). Then [itex]a \in (N^* (x,e) \cap S) \neq \emptyset [/itex] ... which means that [itex]a \in S[/itex]. Since [itex]a \in S[/itex], 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?