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February 3rd, 2013, 09:58 AM   #1
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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?
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February 3rd, 2013, 12:24 PM   #2
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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.
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February 3rd, 2013, 12:33 PM   #3
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Re: Closed set and closed ball

Quote:
Originally Posted by mathman
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.
I think we can create the family of closed balls, so that their intersection will be equal to A.
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February 4th, 2013, 12:49 PM   #4
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Re: Closed set and closed ball

Quote:
Originally Posted by Vasily
Quote:
Originally Posted by mathman
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.
I think we can create the family of closed balls, so that their intersection will be equal to A.
Construct a family of closed balls with intersection consists of two separated points?

Consider 1-d 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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