Let E be a subset of a metric or a topological space Then E

Let E be a subset of a metric or a topological space. Then E is both open and closed if and only if partial differentialE = oslash.

Solution

Let E be a subset of a metric space or a topological space. Then,

E is both open and closed

=> int E=E and E^d=E

Now (int E U dE)=E^d, here dE is the boundary of E.

=> E U dE= E. So dE is either empty or E itself.

But dE=E contradicts int E=E, so dE=null set.

conversely, if dE=null set, then E^d must be E or null set => E is closed.

Again for dE= null set, int E= E => E is open.

So E is both open and closed.