The Closure of a Set in a Topological Space Examples 1

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 set containing $A$.

We will now look at some examples of the closure of a set

Example 1

Consider the topological space $(X, \tau)$ where $\tau$ is the discrete topology on $X$. What is the closure of $A \subseteq X$?

Recall that if $\tau$ is the discrete topology then $\tau = \mathcal P (X)$. Hence every subset of $X$ is open and by extension, every subset of $X$ is closed. Hence, for each $A \subseteq X$ the smallest closed set containing $A$ is $A$ itself!

Therefore we have that:

(1)

\begin{align} \quad \bar{A} = A \end{align}

Example 2

Consider the topological space $(X, \tau)$ where $\tau$ is the indiscrete topology on $X$. What is the closure of $A \subseteq X$?

If $\tau$ is the indiscrete topology then $\tau = \{ \emptyset, X \}$. Furthermore, the only closed sets of $X$ with respect to the indiscrete topology are $\emptyset$ and $X$.

If $A \neq \emptyset$ then the smallest closed set containing $A$ is $X$ itself! Hence:

(2)

\begin{align} \quad \bar{A} = X \end{align}

If $A = \emptyset$ then the smallest closed set containing $A$ is $\emptyset$, so:

(3)

\begin{align} \quad \bar{\emptyset} = \emptyset \end{align}

Example 3

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau = \{ \emptyset \} \cup \{ (-n, n) : n \in \mathbb{Z}, n \geq 1 \}$. What is the closure of $A = \{ 0 \}$? What is the closure of $B = (2, 3)$?

Notice that the open sets of $\mathbb{R}$ with respect to the topology $\tau$ are:

(4)

\begin{align} \quad \tau = \{ \emptyset, (-1, 1), (-2, 2), ..., (-n, n), ..., \mathbb{R} \} \end{align}

Therefore the closed sets of $\mathbb{R}$ with respect to this topology are:

(5)

\begin{align} \quad \mathrm{closed \: sets \: of \:} \mathbb{R} = \{ \emptyset, (-\infty, -1] \cup [1, \infty), (-\infty, -2] \cup [2, \infty) , ..., (-\infty, -n] \cup [n, \infty), ..., \mathbb{R} \} \end{align}

Notice that NONE of these sets except for the whole set $\mathbb{R}$ contain $\{ 0 \}$. Therefore:

(6)

\begin{align} \quad \bar{A} = \bar{\{ 0 \}} = \mathbb{R} \end{align}

Now notice that $(2, 3) \subseteq (-\infty, -2] \cup [2, \infty)$. This is the smallest such closed set, and so:

(7)

\begin{align} \quad \bar{B} = \overline{(2, 3)} = (-\infty, -2] \cup [2, \infty) \end{align}