Boundary of Set A={a} in the Topology {X,{a},{a,b},Ø}

In this exercise, I need to determine the boundary (∂A) of the set $ A = \{a\} $ in $ X = \{a, b, c\} $ with the topology $\{X, \emptyset, \{a\}, \{a, b\}\}$.

In this topology, the open sets are: $ \{a, b, c\}, \{ a \}, \{a, b\}, \emptyset $.

The closed sets are the complements of the open sets: $\{X, \emptyset, \{b,c\}, \{c\}\}$.

The boundary is the intersection between the closure of $ A $ and the closure of the complement of $ A $.

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) $$

The set $ A = \{a\} $ is a singleton. The closure of $A$ is the smallest closed set that contains $A$.

The smallest closed set that contains $\{a\}$ is $X$, which is $\{a, b, c\}$.

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

Next, I need to determine the closure of the complement of $A$.

The complement of $A = \{a\} $ is:

$$ A^c = \{b, c\} $$

The closure of $A^c$ is the smallest closed set that contains $\{b, c\}$.

Since the closed sets in this topology are $\{X, \{c\}, \{b, c\}, \emptyset\}$, the smallest closed set that contains $\{b, c\}$ is $\{b, c\}$.

$$ \text{Cl}(A^c) = \{b, c\} $$

Now I can calculate the intersection of the closures, knowing that $ \text{Cl}(A) = \{a, b, c\} $ and $ \text{Cl}(A^c) = \{b, c\} $

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) $$

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

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

The intersection of these sets is $\{b, c\}$.

So, the boundary of $A = \{a\}$ in $ X = \{a, b, c\} $ with the topology $\{X, \emptyset, \{a\}, \{a, b\}\}$ is $
\partial A = \{b, c\} $.

Alternative Method

I can also solve the problem by considering the boundary of $A$ as the difference between the closure and the interior of $A$,

$$ \partial A = \text{Cl}(A) - \text{int}(A) $$

In this case, I already know the closure of $ A $, so I don't need to recalculate it.

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

Now, I need to determine the interior of $A$, which is $\text{int}(A)$.

The interior of $A$ is the union of all open sets contained within $A$.

In the given topology, the open sets are $\{X, \emptyset, \{a\}, \{a, b\}\}$. Therefore, the only open set contained in $\{a\}$ is $\{a\}$ itself.

$$ \text{int}(A) = \{a\} $$

Now, knowing $ \text{Cl}(A) = \{a, b, c\} $ and $ \text{Int}(A) = \{ a \} $, I can calculate the boundary of $ A $ as the difference between the closure and the interior.

$$ \partial A = \text{Cl}(A) - \text{int}(A) $$

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

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

Therefore, the boundary of $A = \{a\}$ in $X = \{a, b, c\}$ with the topology $\{X, \emptyset, \{a\}, \{a, b\}\}$ is $ \partial A = \{b, c\} $.

And so on.