Boundary of A = {a, b} in X = {a, b, c} Equipped with the Topology {X, Ø, {a}, {a, b}}

In this example, we will determine the boundary \(\partial A\) of the set \(A = \{a, b\}\) in the topological space \(X = \{a, b, c\}\), equipped with the topology \(\{X, \{a\}, \{a, b\}, \emptyset\}\).

The open sets of this topology are

$$ X,\ \{a\},\ \{a, b\},\ \emptyset. $$

The corresponding closed sets, obtained by taking complements in \(X\), are

$$ X,\ \{b, c\},\ \{c\},\ \emptyset. $$

Method 1: Boundary as Closure Minus Interior

The boundary of a set is defined as the difference between its closure and its interior:

$$ \partial A = \operatorname{Cl}(A) \setminus \operatorname{Int}(A). $$

Step 1. Find the closure of \(A\).

The closure of a set is the smallest closed set that contains it.

Since no proper closed subset of \(X\) contains both \(a\) and \(b\), the closure of \(A\) is

$$ \operatorname{Cl}(A) = X = \{a, b, c\}. $$

Step 2. Find the interior of \(A\).

The interior of a set is the largest open set contained in it.

The open subsets of \(A\) are

$$ \emptyset,\ \{a\},\ \{a, b\}. $$

Therefore,

$$ \operatorname{Int}(A) = \{a, b\}. $$

Step 3. Compute the boundary.

Subtract the interior from the closure:

$$ \partial A = \operatorname{Cl}(A) \setminus \operatorname{Int}(A) $$

$$ \partial A = \{a, b, c\} \setminus \{a, b\} $$

$$ \partial A = \{c\}. $$

Therefore, the boundary of \(A\) is

$$ \boxed{\partial A = \{c\}}. $$

Method 2: Boundary as the Intersection of Two Closures

An equivalent definition of the boundary is

$$ \partial A = \operatorname{Cl}(A) \cap \operatorname{Cl}(X \setminus A). $$

We already know that

$$ \operatorname{Cl}(A) = \{a, b, c\}. $$

Now compute the complement of \(A\):

$$ X \setminus A = \{a, b, c\} \setminus \{a, b\} = \{c\}. $$

Since \(\{c\}\) is a closed set, its closure is simply

$$ \operatorname{Cl}(X \setminus A) = \{c\}. $$

Finally, intersect the two closures:

$$ \partial A = \operatorname{Cl}(A) \cap \operatorname{Cl}(X \setminus A) $$

$$ \partial A = \{a, b, c\} \cap \{c\} $$

$$ \partial A = \{c\}. $$

Final Answer

Both methods produce the same result:

$$ \boxed{\partial A = \{c\}}. $$