How is boundary of a set defined? Is boundary of an open set $S$ included in $S$?

Question

How is boundary of a set defined? Is boundary of an open set $S$, included in $S$?

Answer 1

One easy way to understand boundary of a set is the collection of points such that if we draw balls(of any radius) around them,that ball will include some part of set and some part outside of set.(not included in set)

Let's illustrate this with some examples: 1.Consider set $[1,2]$,now it's boundary points are 1 and 2 only.Suppose we claim 1.5 is boundary point, clearly if we draw one ball around it of radius 0.2, this will not contain any part outside of a set.Make same argument for any point outside the set.

2.Now let's consider one open set for your second question say $(2,3)$.The boundary points are 2 and 3.Try on your own to prove that no other point will be boundary point by considering ball around them.This confirms that boundary of an open set is not included in the set.

Answer 2

There are a number of ways to characterize boundaries. I like the algebraic one, expressed in terms of the operation of taking the closure of a set.

Notation: if $A$ is a subset of a topological space, then $\overline{A}$ means its closure, $A^c$ its complement, and $\partial A$ its boundary.

The following facts relating boundaries to open and closed sets are true for any subset $A$ of a topological space:

Theorem: $\partial A = \overline{A} \cap \overline{A^c}$

Theorem: $A$ is closed if and only if $\partial A \subseteq A$

Theorem: $A$ is open if and only if $A \cap \partial A = \varnothing$

Another common definition is through the nearness relation:

Theorem: $x$ is near $A$ if and only if every open set $U$ containing $x$ satisfies $U \cap A \neq \varnothing$

Theorem: $x \in \partial A$ if and only if $x$ is near $A$ and $x$ is near $A^c$

(note that which specific fact is chosen to be the definition is unimportant; it is relatively common for different sources to make different choices)

Answer 3

If $A \subseteq X$ is a subset of a topological space $X$, then $\partial A$ ,the boundary of $A$ in the space $X$ is defined as:

$$\partial A = \{x \in X: \forall O \subseteq X \text{ open }: x \in O \implies O \cap A \neq \emptyset \text{ and } O \cap (X\setminus A) \neq \emptyset\}$$

i.e. the set of points $x$ of $S$ such that every (open) neighbourhood of $x$ intersects both $A$ and its complement. So $x$ is "close" to both $A$ and its complement.

Now, if $A$ is open, for any $x$ in $A$, the $A$ is itself an open set that does not intersect $X\setminus A$, so no point of $A$ can be in the boundary of $A$.

That last observation answers your question: for an open set $A$: $A \cap \partial A = \emptyset$.