Answer to Question #115198 in Differential Geometry | Topology for Sheela John

Question

Prove that the boundary of a subset A of a metric space X is always a closed set

Answer 1

if  A⊆X  then a point x $\in$ X is said to be a boundary point of A if x is contained in the closure of A

 and not in the interior of A

 x∈A∖int(A)

the set of all boundary points of A is called the boundary of A denoted:

∂A=A∖int(A)

To show that ∂A is closed we only need to show that (∂A)^c=X∖∂A is open.

X∖∂A=(A∖∂A)∪(A^c∖∂A)

We will show that A∖∂A and A^c∖∂A are open sets.

Let x∈A∖∂A

 x∈A  and x does not belong to boundary of A

Since x is not on the boundary of A,there exists an open neighborhood K that intersects A^c

 K∩A^c=∅

K⊆A and x∈K⊆A

, so x∈int(A∖∂A)

so A∖∂A  is open.

Let x∈A^c∖∂A

Then x∈A^c and x is not on the boundary of A

Since x is not on the boundary of A there exists an open neighborhood R of x that intersects A

 trivially, i.e., R∩A=∅

 R⊆A^C

 x∈R⊆A^C

 x∈int(A^C∖∂A)

so int(A^C∖∂A)=A^C∖∂A

A^C∖∂A is open.

 (∂A)^C=X∖∂A  is the union of two open sets and so (∂A)^C is open. Therefore ∂A is closed.