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September 19th, 2013, 10:36 AM   #1
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metric space

Hi , i have this problem





Help !
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September 19th, 2013, 11:01 AM   #2
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Re: metric space

I will give a trivial proof and a rigorous proof.
Trivial proof - Every metric space is Hausdorff and in Hausdorff spaces, singletons are closed. Thus is closed in .
Rigorous proof - The singleton is closed if and only if its complement is open. We will verify that the complement of is open. Let . Since then . Let so that the open ball centered at with radius which we'll denote doesn't include either. Thus for any point outside , we have an open set containing that point which does not contain . This is to say that the complement of is open in as required.
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September 20th, 2013, 12:09 AM   #3
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Re: metric space

A subset X of E is closed if every convergent sequence in X converges to a point in X.

For X = {a}, there is only one sequence! Namely: a, a, a ... It converges to a, which is in X. Hence X = {a} is closed.
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September 25th, 2013, 09:47 AM   #4
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Re: metric space

Quote:
Originally Posted by alejandrofigueroa
Hi , i have this problem





Help !
You can't- it's not true! Only subsets of E are open or closed and "a" is a member of E not a subset. I'm not trying to be funny- it is important in proofs to be as precise as possible.

If you mean "{a}" then a set is closed if and only if its complement is open. The complement of {a}, B, is the set of all y such that .
To show that B is open you can show that every point in B is an "interior point". Let y be in B. To show that y is an interior point of B you must show that there exist some neighborhood of y contained in B. Since y is in B, so d(x,y)> 0. Let r= d(x, y). Let , the "ball centered on y with radius r/2". Since d(x, y)= r> r/2, x is NOT in that set so that ball is contained in B, the complement of {a}.
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September 25th, 2013, 10:29 AM   #5
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Re: metric space

Quote:
Originally Posted by HallsofIvy
Let y be in B.
I apologise in advance, but I couldn't resist pointing out that in your proof you didn't cover the case where E is the singleton {a}, hence B, the complement of {a} may be empty.

I'm sure you'll get me back one time, but, as you said, you have to be precise!
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September 29th, 2013, 11:03 AM   #6
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Re: metric space

You are right. I should have said "If y is in B" rather than "let y be in B".
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