Question

Let x be a limit point of a subset A. Let U be an open subset containing x. Prove that x is a limit point of A ∩ U.

Answer 1

To prove this, we need to show that there exists a point y in A ∩ U other than x that is arbitrarily close to x. Since U is an open set containing x, there exists a point y in U such that y ≠ x and y is an element of A. This means x is a limit point of A ∩ U.

Step-by-step explanation:

To prove that x is a limit point of A ∩ U, we need to show that there exists a point y, where y ≠ x, such that y is an element of A ∩ U and y is arbitrarily close to x. Since x is a limit point of A, every open set containing x must contain a point of A other than x. So, since U is an open set containing x, there exists a point y in U such that y ≠ x and y is an element of A. Therefore, y is an element of A ∩ U. Additionally, since U is an open set containing x, there exists a neighborhood around x contained in U. Within this neighborhood, there are infinitely many points of A, which means x is a limit point of A ∩ U.