The Closure of a Set in a Topological Space

Recall from the Accumulation Points of a Set in a Topological Space that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $x \in X$ is called an accumulation point of $A$ if every open neighbourhood $U$ ($U \in \tau$) contains elements of $A$ different from $x$.

We denoted the set of all accumulation points of $A$ as:

(1)

\begin{align} \quad A' = \{ x \in X : x \: \mathrm{is \: an \: accumulation \: point \: of \:} A \} \end{align}

We will now look at another important term known as the closure of $A$.

Definition: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then Closure of $A$ denoted by $\bar{A}$ or $\mathrm{cl} (A)$ is the smallest closed set such that $A \subseteq \bar{A}$.

By $U$ being the "smallest" closed set containing $A$ we mean that if $U$ and $V$ are both closed sets containing $A$ then $A \subseteq U \subseteq V$. Equivalently, the closure of $A$ can be defined to be the the intersection of all closed sets which contain $A$ as a subset.

Proposition 1: Let $(X, \tau)$ be a topological space. a) The closure of the whole set $X$ is $X$, that is, $\overline{X} = X$. b) The closure of the empty set is the empty set, that is, $\overline{\emptyset} = \emptyset$.

Proposition 2 (Idempotency of the Closure of a Set): Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then the closure of the closure of $A$ is equal to the closure of $A$, that is, $\bar{\bar{A}} = \bar{A}$.

• Proof: By definition, $\bar{\bar{A}}$ is the smallest closed set containing $\bar{A}$. But $\bar{A}$ is closed, and so $\bar{\bar{A}} = \bar{A}$. $\blacksquare$

Proposition 3: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then $a \in \bar{A}$ if and only if $U \cap A \neq \emptyset$ for every open neighbourhood $U$ of $a$.

• Proof: $\Rightarrow$ Let $a \in \bar{A}$. Suppose that there exists an open neighbourhood $U$ of $a$ such that $U \cap A = \emptyset$. Then $A \subset X \setminus U$ (note that the inclusion is strict since $Since [[$ U$ is open, $X \setminus U $] is closed. Since [[$ \bar{A}$ is the smallest closed set containing $A$ we see that:

(2)

\begin{align} \quad \bar{A} \subseteq X \setminus U \end{align}

• But this is a contradiction. Observe that $a \in \bar{A}$ but $a \not \in X \setminus U$. Therefore the assumption that such an open neighbourhood $U$ of $a$ exists is false.

• $\Leftarrow$ Suppose that $U \cap A \neq \emptyset$ for every open neighbourhood $U$ of $a$. If $a \not \in \bar{A}$ then $a \in X \setminus \bar{A}$. But $X \setminus \bar{A}$ is open, and so there exists an open neighbourhood $U$ of $a$ such that $U \subseteq X \setminus \bar{A}$.But then $U \cap A = \emptyset$, a contradiction. $\blacksquare$

Let's look at some examples of the closure of a set.

Example 1

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals on $\mathbb{R}$ and let $A = [0, 1)$. The closure of $A$ is $\bar{A} = [0, 1]$. To verify this, we note that $[0, 1]$ is indeed closed because:

(3)

\begin{align} \quad [0, 1]^c = \underbrace{(-\infty, 0)}_{\in \tau} \cup \underbrace{(1, \infty)}_{\in \tau} \in \tau \end{align}

Furthermore, no smaller set contains $A$, so indeed $\bar{A} = [0, 1]$.

Example 2

For another example, consider the set $X = \{ a, b, c, d \}$ and the topology $\tau = \{ \emptyset, \{ b \}, \{ a, b \}, \{ b, c \}, \{a, b, c \}, X \}$ and the set $A = \{ b, d \}$. What is the closure of $A$, i.e., what is $\bar{A}$?

Let's first list out all of the closed sets of $X$. Recall that a set $B$ is said to be closed if $B^c$ is open. So the set of closed sets of $X$ is the set of complements of elements from $\tau$. The closed sets for this topology are therefore:

(4)

\begin{align} \quad \mathrm{Closed \: Sets \: of \:} X = \{ \emptyset, \{a, c, d \}, \{c, d \}, \{a, d \}, \{ d \}, X \} \end{align}

We now need to find the smallest closed set of $X$ such that $A = \{ b, d \} \subseteq X$. In this example, the smallest closed set is the whole set $X$. Therefore $\bar{A} = X$.

Now instead consider the set $B = \{ a \} \subseteq X$. What is $\bar{B}$? Well, the smallest closed set containing $B$ is $\{ a, d \}$ so $\bar{B} = \{ a, d \}$.