Boundary of \(A=\{a,c\}\) in the Topological Space \(X=\{a,b,c\}\)
In this example, we will find the boundary \(\partial A\) of the set \(A=\{a,c\}\) in the topological space \(X=\{a,b,c\}\), equipped with the topology
$$ \tau=\{X,\emptyset,\{a\},\{a,b\}\}. $$
To compute the boundary, we first identify the open and closed sets of the topology.
- Open sets: \(X\), \(\emptyset\), \(\{a\}\), and \(\{a,b\}\).
- Closed sets: \(X\), \(\emptyset\), \(\{b,c\}\), and \(\{c\}\), obtained by taking the complements of the open sets.
We use the standard formula for the boundary:
$$ \partial A=\operatorname{Cl}(A)\setminus\operatorname{Int}(A). $$
Step 1. Find the Closure of \(A\)
The closure of a set is the smallest closed set that contains all of its points.
Since \(A=\{a,c\}\), we look for the smallest closed set containing both \(a\) and \(c\). In this topology, the only closed set that satisfies this condition is the entire space. Therefore,
$$ \operatorname{Cl}(A)=\{a,b,c\}. $$
Step 2. Find the Interior of \(A\)
The interior of a set is the union of all open sets contained in it.
Among the open sets in the topology, the only non-empty open set contained in \(A\) is \(\{a\}\). Hence,
$$ \operatorname{Int}(A)=\{a\}. $$
Step 3. Compute the Boundary
Now subtract the interior from the closure:
$$ \partial A=\operatorname{Cl}(A)\setminus\operatorname{Int}(A) $$
$$ \partial A=\{a,b,c\}\setminus\{a\} $$
$$ \partial A=\{b,c\}. $$
Therefore, the boundary of \(A=\{a,c\}\) is
$$ \boxed{\partial A=\{b,c\}.} $$
Alternative Approach
You can also compute the boundary using the equivalent formula
$$ \partial A=\operatorname{Cl}(A)\cap\operatorname{Cl}(X\setminus A). $$
From the previous steps, we already know that
$$ \operatorname{Cl}(A)=\{a,b,c\}. $$
Next, find the complement of \(A\):
$$ X\setminus A=\{a,b,c\}\setminus\{a,c\}=\{b\}. $$
The smallest closed set containing \(\{b\}\) is
$$ \operatorname{Cl}(X\setminus A)=\{b,c\}. $$
Finally, compute the intersection:
$$ \partial A=\{a,b,c\}\cap\{b,c\} $$
$$ \partial A=\{b,c\}. $$
As expected, this alternative approach leads to the same result:
$$ \boxed{\partial A=\{b,c\}.} $$
