Exercise on the Interior and Closure of a Set in Topology 1

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

Interior of Set A

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

The open sets in the given topology are:

$$ \emptyset $$

$$ \{a\} $$

$$ \{a, b\} $$

$$ X $$

The open sets contained in A are the empty set \(\emptyset\) and {a} itself.

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

Therefore, the interior of A is:

\[ \text{Int}(A) = \{a\} \]

Closure of 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.

Therefore, the closed sets are:

$$ X^c = \emptyset $$

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

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

$$ \emptyset^c = X $$

The smallest closed set that contains A={a} is X={a,b,c}, since none of the other closed sets contain {a}. Therefore, the closure of A is:

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

And that concludes the exercise.