Thread: 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!

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 is a closed set containing and so . That said, since each with is a limit point of (why?) one must have that , can you conclude?


Oh, and it's not as intuitively obvious as one would hope. For example, in a discrete space one has that but for all

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?

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.

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 }?

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 .