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Topology (Boundary points, Interior Points, Closure, etc...) 
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#1
Mar406, 07:20 PM

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. 


#2
Mar406, 07:41 PM

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P: 2,567

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. 


#3
Mar406, 07:53 PM

P: 188

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! 


#4
Mar506, 02:51 AM

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Topology (Boundary points, Interior Points, Closure, etc...)
Defintion of open:
A is open if for any point a in A there is an interval (xe,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 highbrowness. 


#5
Mar506, 09:37 AM

P: 188

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, (xe, 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! 


#6
Mar506, 09:45 AM

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


#7
Mar506, 09:53 AM

P: 188

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 


#8
Mar506, 10:18 AM

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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.



#9
Mar506, 10:34 AM

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." :) 


#10
Mar506, 10:58 AM

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#11
Mar506, 11:14 AM

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. 


#12
Mar506, 11:33 AM

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#13
Mar506, 11:42 AM

P: 188

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.



#14
Mar606, 06:14 PM

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 [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? 


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