Boundary of the Set \(A=\{c\}\) in \(X=\{a,b,c\}\) with Topology \(\{X,\emptyset,\{a\},\{a,b\}\}\)

To find the boundary of the set \(A=\{c\}\), we first determine its closure and its interior in the topological space \(X=\{a,b,c\}\) equipped with the topology

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

Once these two sets have been identified, the boundary follows directly from the definition:

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

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

The open sets of the topology are

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

so the closed sets are their complements:

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

Since \(A=\{c\}\) is already a closed set, its closure is the set itself:

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

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

The interior of a set is its largest open subset.

Neither \(\{a\}\) nor \(\{a,b\}\) is contained in \(\{c\}\), so the only open subset of \(A\) is the empty set. Therefore,

$$ \operatorname{Int}(A)=\emptyset. $$

Step 3. Compute the Boundary

Substituting the closure and the interior into the definition gives

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

$$ \partial A=\{c\}\setminus\emptyset $$

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

Hence, the boundary of the set is

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

Alternative Method

You can also compute the boundary using the equivalent formula

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

We already know that

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

The complement of \(A\) is

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

The smallest closed set containing \(\{a,b\}\) is the whole space \(X\), so

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

Now compute the intersection:

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

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

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

This confirms the previous result:

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