The interior of a set is the largest open set inside that set. The interior of the set \(A\) is written \(A^\circ.\)
The closure of a set is the smallest closed set containing that set. The closure of the set \(A\) is written \(\overline{A}.\)
Proposition: \(\overline{A} = A \cup A'.\)
Proof: We will show \(\overline{A} \subset A \cup A'\) and \(A \cup A' \subset \overline{A}.\)
First, let \(x \in \overline{A}.\) If \(x \in A,\) we are done. If \(x \not\in A,\) then by way of contradiction suppose \(x \not\in A'.\) Then there is an open interval \(I\) containing \(x\) such that \(I \cap A = \emptyset.\) Since \(I\) is open, \(I^C\) is closed, so \(I^C \cap \overline{A}\) is closed. However, \(A \subset I^C \cap \overline{A}\) and \(I^C \cap \overline{A}\) is a proper subset of \(\overline{A},\) which is a contradiction. So, \(x \in A'.\)
To go the other direction, assume \(x \in A \cup A'.\) Again, if \(x \in A\) then it is automatic that \(x \in \overline{A}\) since \(A \subset \overline{A}.\) If \(x \not\in A,\) then \(x \in A'.\) Suppose by way of contradiction that \(x \not\in \overline{A}.\) Since \(\overline{A}^C\) is open, there must exist an open interval \(I\) such that \(x \in I\) and \(I \subset \overline{A}^C.\) Since \(x \in A',\) there must exist a point \(a \in A\) such that \(a \in I.\) However, if \(a \in I,\) then \(a \not\in \overline{A}.\) This is a contradiction to the definition of \(\overline{A}.\) So, \(x \in \overline{A}.\)