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 .

Anuj holds professional certifications in Google Cloud, AWS as well as certifications in Docker and App Performance Tools such as New Relic. He specializes in Cloud Security, Data Encryption and Container Technologies.

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