Exercise on the Interior and Closure of a Set in Topology 4

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

Interior of the Set A
Closure of the Set A

Interior of the Set A

The interior of $A = \{a, b\} $ is the union of all open sets contained within $A$.

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

So, the open sets contained within $A = \{a, b\}$ are:

$$ \{a\} $$

$$ \{a, b\} $$

The union of these sets is $\{a, b\}$.

$$ \text{Int}(A) = \{a\} \cup \{a, b\} = \{a, b\} $$

Therefore, the interior of $A = \{a, b\}$ is always the set $\{a, b\}$ because it is open.

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

Closure of the Set A

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

Closed sets are the complements of the open sets.

Given that in this topology the open sets are $X$, $\emptyset$, $\{a\}$, and $\{a, b\}$, the closed sets are:

$$ X^c = \emptyset $$

$$ \emptyset^c = X $$

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

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

Only the closed set $X = \{a, b, c\}$ contains $A = \{a, b\}$.

Therefore, the closure of $A$ is the set $X$.

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

And that's it.