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February 3rd, 2013, 09:58 AM  #1 
Senior Member Joined: Apr 2012 Posts: 106 Thanks: 0  Closed set and closed ball
Here is the claim: If set A is closed, then there exists a family of sets D, whose elements are closed balls, so that Here is another claim: If set A is open, then there exist family D, whose elements are open balls, to that Here is a try: i take X that is in . X is also in open D_n so it's neighbourghood is also in D_n. D_n is a part of , therefore neighbourghood of X is also in . Which means that open balls form an open set. I am not sure if this is a right way of thinking. What if it's reverse. So what happens if I take that x is in A? 
February 3rd, 2013, 12:24 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,527 Thanks: 588  Re: Closed set and closed ball
The statement about open sets is correct and your proof is valid. The closed set statement is obviously false. Consider a closed set consisting of just two points. An intersection of closed balls will be connected or empty. 
February 3rd, 2013, 12:33 PM  #3  
Senior Member Joined: Apr 2012 Posts: 106 Thanks: 0  Re: Closed set and closed ball Quote:
 
February 4th, 2013, 12:49 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,527 Thanks: 588  Re: Closed set and closed ball Quote:
Consider 1d case. Closed balls are closed intervals. The intersection of two intervals is either empty or an interval. You can never get two separate pieces.  

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ball, closed, set 
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