The Closure of a Set Equals the Union of the Set and its Acc. Points.

# The Closure of a Set Equals the Union of the Set and its Accumulation Points

Recall from The Closure of a Set in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then the closure of $A$ is the smallest closed subset containing $A$ denoted $\bar{A}$.

We will now look at a very nice theorem which relates the set $A$, the set of accumulation points $A'$, and the closure $\bar{A}$.

Theorem: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then $\bar{A} = A \cup A'$. |

**Proof:**Let $x \in \bar{A}$. Then $x$ is contained in the smallest closed set containing $A$. Hence $x \in A$ or $x \in \bar{A} \setminus A$. If $x \in A$ then $x \in A \cup A'$. If $x \not \in A$, then $x \in \bar{A} \setminus A$.

- Since $\mathrm{int} (A) \subseteq A$ we see that if $x \not \in A$ then $x \not \in \mathrm{int} (A)$. Therefore, there exists no open neighbourhood $U \in \tau$ with $x \in U$ such that $x \in U \subseteq A$. Hence for all $U \in \tau$ with $x \in U$ we have that $A \cap U = A \cap U \setminus \{ x \} \neq \emptyset$ so $x \in A'$. Hence $x \in A \cup A'$.

- So, $\bar{A} \subseteq A \cup A'$.

- (Alternatively since a set is closed if and only if it contains all of its accumulation points, we see that $A \cup A'$ is closed as it contains all of its accumulation points. Since $\bar{A}$ is the smallest closed set containing $A$ we have that $\bar{A} \subseteq A \cup A'$.)

- $\Rightarrow$ Now suppose that $A \cup A' \not \subseteq \bar{A}$. Then there exists an $x \in A \cup A'$ such that $x \not \in \bar{A}$. If $x \in A$ then since $A \subseteq \bar{A}$ we have that $x \in \bar{A}$ which is a contradiction. Hence $x \in A' \setminus A$. So then $x$ is an accumulation point of $A$, so for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$. Since $A \subseteq \bar{A}$ we have that then for all $U \in \tau$ with $x \in U$ that $\bar{A} \cap U \setminus \{ x \} \neq \emptyset$. Therefore $x$ is an accumulation point of $\bar{A}$. But $\bar{A}$ is closed and by definition contains all of its accumulation points, so $x \in \bar{A}$, a contradiction.

- Hence the assumption that $A \cup A' \not \subseteq \bar{A}$ was false. Therefore $A \cup A' \subseteq \bar{A}$.

- We conclude that then $\bar{A} = A \cup A'$.