Boundary of the Set \(A=\{a\}\) in the Topology \(\tau=\{X,\emptyset,\{a\},\{a,b\}\}\)
In this example, we will find the boundary \(\partial A\) of the set \(A=\{a\}\) in the topological space \(X=\{a,b,c\}\) equipped with the topology
$$ \tau=\{X,\emptyset,\{a\},\{a,b\}\}. $$
The open sets in this topology are
$$ X,\ \emptyset,\ \{a\},\ \{a,b\}. $$
The corresponding closed sets, obtained by taking complements in \(X\), are
$$ X,\ \emptyset,\ \{b,c\},\ \{c\}. $$
To find the boundary of \(A\), we use the definition
$$ \partial A=\operatorname{Cl}(A)\cap\operatorname{Cl}(X\setminus A), $$
where \(\operatorname{Cl}(A)\) denotes the closure of \(A\).
Step 1: Find the Closure of \(A\)
The closure of a set is the smallest closed set containing it.
Since \(A=\{a\}\), we look for the smallest closed set that contains the point \(a\). From the list of closed sets, the only one is the entire space \(X\).
Therefore,
$$ \operatorname{Cl}(A)=X=\{a,b,c\}. $$
Step 2: Find the Closure of the Complement of \(A\)
The complement of \(A\) is
$$ X\setminus A=\{b,c\}. $$
Next, we determine the smallest closed set containing \(\{b,c\}\).
Since \(\{b,c\}\) is already a closed set, its closure is simply
$$ \operatorname{Cl}(X\setminus A)=\{b,c\}. $$
Step 3: Compute the Boundary
Now apply the definition of the boundary:
$$ \partial A=\operatorname{Cl}(A)\cap\operatorname{Cl}(X\setminus A). $$
Substituting the two closures gives
$$ \partial A=\{a,b,c\}\cap\{b,c\}. $$
Hence,
$$ \partial A=\{b,c\}. $$
Therefore, the boundary of the set \(A=\{a\}\) is
$$ \boxed{\partial A=\{b,c\}.} $$
Alternative Method
You can also compute the boundary using the identity
$$ \partial A=\operatorname{Cl}(A)\setminus\operatorname{Int}(A), $$
where \(\operatorname{Int}(A)\) denotes the interior of \(A\).
We already know that
$$ \operatorname{Cl}(A)=\{a,b,c\}. $$
Now compute the interior of \(A\).
The interior of a set is the largest open set contained in it.
Because \(\{a\}\) is itself an open set in this topology, we have
$$ \operatorname{Int}(A)=\{a\}. $$
Applying the formula gives
$$ \partial A=\{a,b,c\}\setminus\{a\}. $$
Therefore,
$$ \partial A=\{b,c\}. $$
As expected, both methods lead to the same result:
$$ \boxed{\partial A=\{b,c\}.} $$
