Q 66. - Calculus

Question

Let S be a subset of $R^2$ or $R^3$. Prove that a set S is closed if and only if $\partial S\subseteq S$. 

Step-by-Step Solution

Answer

It is proved that " the set is closed if and only if $\partial S\subseteq S$.

Step 1. Given information.

We have given S is a subset of $R^2$ or $R^3$.

Step 2: Prove the given statement.

The objective is to prove that the set is closed if and only if $\partial S\subseteq S$

 Assume an element x' such that $X\in S$, where $S$ is a closed set.

By the definition of a closed set, if $S$ is a closed set, then its complement $S^e$ is an open set. 
This means that for element 'X' there exists such an open disk or ball D, such that D would include or extend beyond the boundary of S. 

Thus, the set $D\cap S^e$or $\partial S\cap S$ is non-empty.
From these statements, it is clear that $\partial S\subseteq S$

In another case, consider S is an open set.

 A set is said to be open if for every element of it, there exists an open disk or ball D, such that$x\in D\subseteq S$
This would mean that 'x' does not belong to the boundary of S. 
$x\notin \partial S$
Thus, there is no common element between S and $\partial S$ This means that as $\partial S\notin S$

Combining the two proofs, it is proved that, "that the set is closed if and only if $\partial S\subseteq S$