September 19th, 2013, 09:36 AM  #1 
Newbie Joined: Sep 2013 Posts: 15 Thanks: 0  metric space
Hi , i have this problem Help ! 
September 19th, 2013, 10:01 AM  #2 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  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. 
September 19th, 2013, 11:09 PM  #3 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115  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. 
September 25th, 2013, 08:47 AM  #4  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: metric space Quote:
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}.  
September 25th, 2013, 09:29 AM  #5  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115  Re: metric space Quote:
I'm sure you'll get me back one time, but, as you said, you have to be precise!  
September 29th, 2013, 10:03 AM  #6 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  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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