Basic Theorems Regarding the Closure of Sets in a Topological Space

# Basic Theorems Regarding the Closure of Sets in a Topological Space

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

We also proved the very important fact that the closure of $A$ is equal to the union of $A$ with its set of accumulation points $A'$, that is, $\bar{A} = A \cup A'$.

We will now look at some basic theorems regarding the closure of sets in a topological space.

 Theorem 1: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. If $A \subseteq B$ then $\bar{A} \subseteq \bar{B}$.
• Proof: Suppose that $A \subseteq B$. Let $x \in \bar{A}$. Since $\bar{A} = A \cup A'$ we have that $x \in A \cup A'$. If $x \in A$ then since $A \subseteq B$ we have that $x \in B$ and $B \subseteq \bar{B}$ so $x \in \bar{B}$.
• If $x \in A'$ then we've already proven that $A' \subseteq B'$ and $B' \subseteq \bar{B}$ (since $\bar{B} = B \cup B'$) so $x \in \bar{B}$.
• In both cases we see that $x \in \bar{A}$ implies that $x \in \bar{B}$, so $\bar{A} \subseteq \bar{B}$. $\blacksquare$

In general if $\bar{A} \subseteq \bar{B}$ then we cannot conclude that $A \subseteq B$. For example, consider the set $X = \{ a, b, c, d \}$ and the topology $\tau = \{ \emptyset, \{a \}, \{ b \}, \{a , b \}, \{a, b, c \}, X \}$. Then the closed sets are:

(1)
\begin{align} \quad \mathrm{closed \: sets} = \{ \emptyset, \{ d \}, \{c, d \}, \{b, c, d \}, \{a, c, d \}, X \} \end{align}

Let $A = \{c, d \}$ and $B = \{ a, c \}$. Then $\bar{A} = \{ c, d \}$ and $\bar{B} = \{ a, c, d \}$, so $\bar{A} \subseteq \bar{B}$. However $A \not \subseteq B$!

 Theorem 2: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\bar{A} \cup \bar{B} = \overline{A \cup B}$.
• Proof: Let $x \in \bar{A} \cup \bar{B}$. Then:
(2)
\begin{align} \quad x \in (A \cup A') \cup (B \cup B') = (A \cup B) \cup (A' \cup B') = \overline{A \cup B} \end{align}
• So $\bar{A} \cup \bar{B} \subseteq \overline{A \cup B}$.
• Similarly, if $x \in \overline{A \cup B}$ then:
(3)
\begin{align} \quad x \in (A \cup B) \cup (A \cup B)' = (A \cup B) \cup (A' \cup B') = (A \cup A') \cup (B \cup B') = \bar{A} \cup \bar{B} \end{align}
• So $\overline{A \cup B} \subseteq \bar{A} \cup \bar{B}$.
• Hence we conclude that $\bar{A} \cup \bar{B} = \overline{A \cup B}$. $\blacksquare$
 Theorem 3: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\bar{A} \cap \bar{B} \supseteq \overline{A \cap B}$.
• Proof: Let $x \in \overline{A \cap B}$. Then $x \in (A \cap B) \cup (A \cap B)' = (A \cap B) \cap (A' \cap B')$. Let $C = A' \cap B'$. Then:
(4)
\begin{align} \quad x \in (A \cap B) \cup (A' \cap B') & = (A \cap B) \cup C \\ & = C \cup (A \cap B) \\ & = (C \cup A) \cap (C \cup B) \\ & = (A \cup C) \cap (B \cup C) \\ & = (A \cup [A' \cap B']) \cap (B \cup [A' \cap B']) \\ & = ([A \cup A'] \cap [A \cup B']) \cap ([B \cup A'] \cap [B \cup B']) \\ & = ([\underbrace{A \cup A'}_{=\bar{A}}] \cap [\underbrace{B \cup B'}_{=\bar{B}}]) \cap ([A \cup B'] \cap [B \cup A']) \\ & = (\bar{A} \cap \bar{B}) \cap ([A \cup B'] \cap [B \cup A']) \end{align}
• So $x \in (\bar{A} \cap \bar{B}) \cap ([A \cup B'] \cap [B \cup A'])$, i.e., $x \in \bar{A} \cap \bar{B}$, so $\bar{A} \cap \bar{B} \supseteq \overline{A \cap B}$. $\blacksquare$

Important Note!

Note that in general $\bar{A} \cap \bar{B} \not \subseteq \overline{A \cap B}$. For example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$ of open intervals.

Consider the set $A = [0, 1)$ and $B = (1, 2]$. Then $\bar{A} = [0, 1]$ and $\bar{B} = [1, 2]$, and so:

(5)
\begin{align} \quad \bar{A} \cap \bar{B} = \{ 1 \} \end{align}

However $A \cap B = \emptyset$ and so:

(6)
\begin{align} \quad \bar{A \cap B} = \bar{\emptyset} = \emptyset \end{align}

Clearly $\bar{A} \cap \bar{B} \not \subseteq \bar{A \cap B}$!

For another example, consider the set $X = \{a, b, c, d \}$ and let $\tau = \{ \emptyset, \{a \}, \{b, c \}, \{a, b, c \}, X \}$. The closed sets are therefore:

(7)
\begin{align} \quad \mathrm{closed \: sets} = \{ \emptyset, \{ d \}, \{ a, d \}, \{b, c, d \}, X \} \end{align}

Consider the sets $A = \{ a, c \}$ and $B = \{ a, b \}$. Then $\bar{A} = X$ and $\bar{B} = X$ so $\bar{A} \cap \bar{B} = X$.

However $A \cap B = \{ a \}$ and $\overline{A \cap B} = \{ a, d \}$.

Clearly $\bar{A} \cap \bar{B} = X \not \subseteq \{a, d \} = \overline{A \cap B}$.