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

  • Thread starter rad0786
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  • #1
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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.
 

Answers and Replies

  • #2
StatusX
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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
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StatusX said:
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.

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
matt grime
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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.
 
  • #5
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matt grime said:
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!
 
  • #6
matt grime
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rad0786 said:
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.
 
  • #7
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matt grime said:
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


matt grime said:
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.
 
  • #8
matt grime
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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
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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
shmoe
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rad0786 said:
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.
 
  • #11
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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
shmoe
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rad0786 said:
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
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}).
 
  • #13
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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
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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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