*Let A be an open set. Show that if a finite number of points are removed from A, the remaining set is still open. Is the same true if a countable number of points are removed? *

### Solution

If we order the points removed from A, x1 < x2 < … < xk we can write A − {x1, x2, …, xk} = A ∩ ((−∞, x1) ∪ (x1, x2) ∪ (x2, x3) ∪ …(xk−1, xk) ∪ (xk, ∞)) B := (−∞, x1) ∪ (x1, x2) ∪ (x2, x3) ∪ …(xk−1, xk) ∪ (xk, ∞) is a union of open sets, hence open. A ∩ B is a finite intersection of open sets, hence it is also open.

The statement is not true for countable number of points. Consider R − Q. This is not open, since given any irrational point x, by the density of rational, we can find for every n a rational y ∈ (x − 1/n, x + 1/n), hence we couldn’t find a neighborhood around x completely in R − Q .