Let X d be a metric space For any A X we define the interior

Let (X, d) be a metric space. For any A X, we define the interior of A in (X, d) as the set A degree:= the union of all open balls in (X, d) that are fully contained in A. Show that A degree is the largest open set (in (X, d)) that is contained in A. (I.e., show that, if U open in (X, d), and U A, then U A degree).

Solution

Given ,

Let (X,d) be a metric space. For any A X, A⁰ = The union of all open balls in (X,d) that are fully contained in A.

We have to prove : A⁰ is the largest open set that is contained in A.

Proof :- If U is open in (X,d) and U A.

If p is an interior point of U,then there is some neighbourhood N of p with NU.

Since U A , N A.

Which shows that p is interior point of A.

Thus U⁰ A^{0...................................}(1)

If U is open, all of its points are interior points, so that U U⁰ .

Also, the set of interior points of U is a subset of the set of points of U, so that U⁰ U. Thus U⁰ = U............(2)

From (1) and (2)

We get U A⁰.

Hence proved.