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December 14th, 2016, 08:41 AM   #1
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limit point

A is a group,A' is the group that containts all the limit point of A.
i have to prove A is closed.
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December 14th, 2016, 02:53 PM   #2
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Maybe I'm missing something. But in set theory a set is closed if it contains all its limit points.
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December 15th, 2016, 12:55 AM   #3
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Maybe I'm missing something. But in set theory a set is closed if it contains all its limit points.
i need to prove that lets say we have group A,some group. lets say we have group A' which containts all the limit point of A,i have to prove that A' is closed.
the definition that you said is right,but i have to prove that A' is doing so.
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December 15th, 2016, 01:05 AM   #4
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i need to prove that lets say we have group A,some group. lets say we have group A' which containts all the limit point of A,i have to prove that A' is closed. i try to prove like this,tell me if its enough. lets say A' isnt closed,which means it has X that doesnt belong to A' and this X is limit point in A'.moreover, we have infinity number in A' (x+e,x-e)(as limit point definition).that means we have limit points of A (cause A' contains A imiit points) in (x-e,x+e).But this X isnt limit point in A(cause he doesnt belong to A') which mean we have finite num in (x+e,x-e) which means we dont have limit points in this part. and thats is contradiction.
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December 15th, 2016, 02:08 PM   #5
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I don't understand why you are working with groups, since it is a set theory question.

The only question in my mind is A' a subset of A?

If A' is a subset of A, then any limit point of A' is a limit point of A, and therefore is in A' and A' is closed.
If A' is not a subset of A, then it could have a limit point that is not in A, and therefore may not be closed.
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December 19th, 2016, 06:00 PM   #6
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To talk about "limit points" you have to have a topology. Are you talking about topological groups?
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December 19th, 2016, 06:18 PM   #7
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The only question in my mind is A' a subset of A?
Surely A' is a superset of A since every element of A can be a limit point.
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December 29th, 2016, 11:51 AM   #8
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Can be, but isn't necessarily. If A is $\displaystyle (0, 1)\cup {2}$ then every point of [0, 1] is a limit point of A but 2 is not. A'= [0, 1] which is NOT a superset of A.
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December 29th, 2016, 12:43 PM   #9
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I'm not sure of the definition of a limit point in this context, but the limit of the sequence $(2,2,2,\ldots)$ is $2$. So my informal definition of a limit point would include $2$ for your set.
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January 5th, 2017, 12:10 AM   #10
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Originally Posted by v8archie View Post
I'm not sure of the definition of a limit point in this context, but the limit of the sequence $(2,2,2,\ldots)$ is $2$. So my informal definition of a limit point would include $2$ for your set.
The definition of a limit point specifically excludes this. Typically one defines a limit point as follows:

If $X$ is a topological space, $A \subset X$ and $x \in X$, then $x$ is a limit point of $A$ if for every neighborhood (in the $X$ topology), $N$ of $x$ we have $N \setminus \{x\} \cap A \neq \emptyset$.
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