Question

Prove that the closure of A intersection B is a subset of the closure of A intersection closure of B. Give an example to show that equality does not hold.

Answer 1

To prove that the closure of A intersection B is a subset of the closure of A intersection closure of B, we need to show that every element in the closure of A intersection B is also in the closure of A intersection closure of B. An example to show that equality does not hold is when A is the closed interval [0, 1], B is the set (1, 2), and C is the closed interval [1, 2].

Step-by-step explanation:

To prove that the closure of A intersection B is a subset of the closure of A intersection closure of B, we need to show that every element in the closure of A intersection B is also in the closure of A intersection closure of B.

Let x be an element in the closure of A intersection B. This means that for any open set U containing x, U intersects A and B. Therefore, U also intersects closure of A and closure of B. Hence, x is in the closure of A intersection closure of B.

An example to show that equality does not hold is when A is the closed interval [0, 1], B is the set (1, 2), and C is the closed interval [1, 2]. Here, the closure of A intersection B is the closed interval [1, 1], but the closure of A intersection closure of B is the closed interval [0, 2]. Therefore, the closure of A intersection B is a proper subset of the closure of A intersection closure of B.