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!
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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 $\displaystyle RHS$ is a closed set containing $\displaystyle B_1(a)$ and so $\displaystyle \overline{B_1(a)}\subseteq RHS$. That said, since each $\displaystyle x$ with $\displaystyle \|x-a\|=1$ is a limit point of $\displaystyle B_1(a)$ (why?) one must have that $\displaystyle RHS\subseteq\overline{B_1(a)}$, can you conclude?Quote:
Originally Posted by AKTilted
Oh, and it's not as intuitively obvious as one would hope. For example, in a discrete space $\displaystyle (X,d)$ one has that $\displaystyle \overline{B_1(x)}=B_1(x)$ but $\displaystyle \left\{y\in X:d(x,y)=1\right\}=X$ for all $\displaystyle x\in X$
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.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 $\displaystyle \overline{B_1(a)}\subseteq\left\{x\in\mathbb{R}^n: \|x-a\|\leqslant 1\right\}$.Quote:
Originally Posted by AKTilted
