# Closure Proof

• Sep 18th 2011, 07:32 PM
AKTilted
Closure Proof
Hi all,
I'm just wondering if you could show me how to prove this statement. Obviously, it's intuitively clear but I just can't get a proof working for it:

Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }

Thanks a lot!
• Sep 18th 2011, 07:53 PM
Drexel28
Re: Closure Proof
Quote:

Originally Posted by AKTilted
Hi all,
I'm just wondering if you could show me how to prove this statement. Obviously, it's intuitively clear but I just can't get a proof working for it:

Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }

Thanks a lot!

You know that $RHS$ is a closed set containing $B_1(a)$ and so $\overline{B_1(a)}\subseteq RHS$. That said, since each $x$ with $\|x-a\|=1$ is a limit point of $B_1(a)$ (why?) one must have that $RHS\subseteq\overline{B_1(a)}$, can you conclude?

Oh, and it's not as intuitively obvious as one would hope. For example, in a discrete space $(X,d)$ one has that $\overline{B_1(x)}=B_1(x)$ but $\left\{y\in X:d(x,y)=1\right\}=X$ for all $x\in X$
• Sep 18th 2011, 08:14 PM
AKTilted
Re: Closure Proof
Does this line of logic work?

If I assume that the closure of the ball, B(1,a) = { x in R^n: |x-a| <= 1 } and then go to prove that this is closed?
• Sep 18th 2011, 08:30 PM
Drexel28
Re: Closure Proof
Quote:

Originally Posted by AKTilted
Does this line of logic work?

If I assume that the closure of the ball, B(1,a) = { x in R^n: |x-a| <= 1 } and then go to prove that this is closed?

I'm sorry, I can't understand what you mean there.
• Sep 18th 2011, 08:43 PM
AKTilted
Re: Closure Proof
Basically this:

If I can prove that { x in R^n: |x-a| <= 1 } is a closed set, can I use this to state that Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }?
• Sep 18th 2011, 08:57 PM
Drexel28
Re: Closure Proof
Quote:

Originally Posted by AKTilted
Basically this:

If I can prove that { x in R^n: |x-a| <= 1 } is a closed set, can I use this to state that Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }?

No..you can use it to state that $\overline{B_1(a)}\subseteq\left\{x\in\mathbb{R}^n: \|x-a\|\leqslant 1\right\}$.