December 14th, 2016, 08:41 AM  #1 
Newbie Joined: Nov 2016 From: Uk Posts: 11 Thanks: 0  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. 
December 14th, 2016, 02:53 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558 
Maybe I'm missing something. But in set theory a set is closed if it contains all its limit points.

December 15th, 2016, 12:55 AM  #3  
Newbie Joined: Nov 2016 From: Uk Posts: 11 Thanks: 0  Quote:
the definition that you said is right,but i have to prove that A' is doing so.  
December 15th, 2016, 01:05 AM  #4 
Newbie Joined: Nov 2016 From: Uk Posts: 11 Thanks: 0 
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,xe)(as limit point definition).that means we have limit points of A (cause A' contains A imiit points) in (xe,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,xe) which means we dont have limit points in this part. and thats is contradiction.

December 15th, 2016, 02:08 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558 
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. 
December 19th, 2016, 06:00 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,921 Thanks: 785 
To talk about "limit points" you have to have a topology. Are you talking about topological groups?

December 19th, 2016, 06:18 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra  
December 29th, 2016, 11:51 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,921 Thanks: 785  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.

December 29th, 2016, 12:43 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,117 Thanks: 2369 Math Focus: Mainly analysis and algebra 
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.

January 5th, 2017, 12:10 AM  #10  
Senior Member Joined: Sep 2016 From: USA Posts: 241 Thanks: 126 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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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