The Closure of a Set in a Topological Space

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:

\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 $\bar{A}$ 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 $A \subseteq U \subseteq V$.

For example, 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:

\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 $\bar{A} = [0, 1]$.

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:

\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 $\bar{B} = \{ a, d \}$.

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