The Closure of a Set in a Topological Space
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# 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}$is the smallest closed set containing$Awe see that: (2) \begin{align} \quad \bar{A} \subseteq X \setminus U \end{align} • But this is a contradiction. Observe thata \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 containsA$, so indeed$\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 ofX$such that$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 \}\$.

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