Real Analysis?

Question

I'm having trouble with the following question.

Suppose A is a closed set and let x be in A. Let B = A / {x} so that B is the set A with the element x removed. Under what conditions is B closed?

So far, I know that x cannot be a limit point of A, since closed sets contain all of their limit points; but I'm having trouble coming up with any further condtions. Thanks.

Answer 1

That's the only condition. B is closed iff x is not a limit point of B.

Proof: First, suppose x is not a limit point of B. Let y be any limit point of B. Then y is a limit point of A (since B⊆A), so because A is closed, y∈A. However, since x is not a limit point of B, y≠x, so y∈A\{x} = B. Since B contains all of its limit points, B is closed. This takes care of the if statement.

For the only if, suppose x is a limit point of A. Then if N is an arbitrary neighborhood of x, ∃y∈N∩A s.t. y≠x. But since y≠x, this means y∈N∩(A\{x}) = N∩B. So for every neighborhood of x, N∩B contains the aforementioned point y≠x, so x is a limit point of B. But since x is not in B, this means B is not closed. Q.E.D.

littleJaina, 1 is not a limit point of ℕ. To be a limit point, every neighborhood of 1 would have to contain a natural number other than 1 itself, but (1/2, 3/2) is a neighborhood of 1 containing no natural numbers other than 1. Indeed, ℕ has no limit points in ℝ.

Answer 2

Yikes! You're bringing back nightmares!!!! Get a study group man, that's the only way I passed that class. Like five of us used to meet twice a weak for several hours to pound out homework and study for tests. Good Luck!!!

However, I think you might need to re-examine whether or not x can be a limit point of A. If I remember correctly, N is a closed set. However, if you take 1 out of N, N/{1} can still be closed. However, N is discrete. Maybe I'm remembering wrong. Good Luck!!!