Exercise on the Interior and Closure of a Set in Topology 3

In this exercise, we need to determine the interior $\text{Int}(A)$ and the closure $\text{Cl}(A)$ of the set $A = \{b\}$ in $X = \{a, b, c\}$ with the topology $\{X, \emptyset, \{a\}, \{a, b\}\}$.

Interior of the Set A
Closure of the Set A

Interior of the Set A

The interior of A={b} is the union of all open sets contained in A.

The open sets in this topology are $X$, $\emptyset$, $\{a\}$, and $\{a, b\}$.

Among these, only $\emptyset$ is contained in $\{b\}$.

Therefore, the interior of A={b} is the empty set, Ø.

$$ \text{Int}(A) = \emptyset $$

Closure of the Set A

The closure of $A$ is the intersection of all closed sets containing A.

Closed sets are the complements of open sets.

In this topology, the open sets are $X$, $\emptyset$, $\{a\}$, and $\{a, b\}$, so the closed sets are:

$X^c = \emptyset$
$\emptyset^c = X = \{a, b, c\}$
$\{a\}^c = \{b, c\}$
$\{a, b\}^c = \{c\}$

The closed sets containing A={b} are X={a,b,c} and {b, c}. The intersection of these sets is {b,c}.

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

Therefore, the closure of the set A={b} is the set {b,c}.

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

And that's it.