Question

Let (X,d) be a metric space, and A ⊂ X (not necessarily a

subspace, just a subset). Show that:
= V is open in X and V ⊂ A

Answer 1

To prove V corresponds to open sets in the subspace topology on A, we utilize the definitions of open sets in a metric space and the subspace topology. If V is open in X and contained in A, it is open in A's subspace topology; conversely, sets open in the subspace can be expressed as intersections with A, implying they stem from open sets in X.

Step-by-step explanation:

To show that V is open in X and V ⊆ A is equal to the collection of subsets of A that are open in the subspace topology on A induced by the metric space (X,d), we need to understand the concept of a subspace topology. In a metric space (X,d), a subset V is open in X if for every point x in V, there exists an ε > 0 such that the ε-ball around x is entirely contained within V. In terms of the subspace topology on A, a set U is open in A if U = V ∩ A for some open set V in X.

We prove the statement by considering two parts:

1. If V is open in X and V ⊆ A, then V is open in the subspace topology on A. Since V is open in X, for every point x in V, there is an ε-ball contained in V. As V ⊆ A, the intersection of this ε-ball with A is still contained in V. Hence, V is open in the subspace topology on A because it can be written as the intersection of an open set in X with A.
2. If U is open in the subspace topology on A, then U can be written as U = V ∩ A for some open set V in X. By the definition of the subspace topology, this V must be open in X, and since V ∩ A = U and U ⊆ A, we have V ⊆ A as well. Thus, U is a part of the set V is open in X and V ⊆ A.

Hence, we have shown that V is open in X and V ⊆ A is the collection of subsets of A that are open in the subspace topology on A induced by (X,d).