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 September 19th, 2013, 09:36 AM #1 Newbie   Joined: Sep 2013 Posts: 15 Thanks: 0 metric space Hi , i have this problem $If \ (E, d) \ a \ metric \ space \ and \ a \ \in \ E\ .\ Prove \ that \ "\ a\ " \ is \ closed.$ 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 $\{a\}$ is closed in $(E, d)$. Rigorous proof - The singleton $\{a\}$ is closed if and only if its complement is open. We will verify that the complement of $\{a\}$ is open. Let $b \in (E, d) \ ; \ b \neq a$. Since $b \neq a$ then $d(b, a) > 0$. Let $d(b, a)= \epsilon$ so that the open ball centered at $b$ with radius $\frac{\epsilon}{2}$ which we'll denote $B(b ,\frac{\epsilon}{2})$ doesn't include $\{a\}$ either. Thus for any point outside $\{a\}$, we have an open set containing that point which does not contain $\{a\}$. This is to say that the complement of $\{a\}$ is open in $(E, d)$ as required.
 September 19th, 2013, 11:09 PM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 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
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Re: metric space

Quote:
 Originally Posted by alejandrofigueroa Hi , i have this problem $If \ (E, d) \ a \ metric \ space \ and \ a \ \in \ E\ .\ Prove \ that \ "\ a\ " \ is \ closed.$ 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 $y\ne x$.
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, $y\ne x$ so d(x,y)> 0. Let r= d(x, y). Let $B_r(y)= \{z| d(y, z)< r/2\}=$, 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
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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!

 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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