Basic Thms. Regarding the Accumulation Points of Sets in a Topo. Sp.

# Basic Theorems Regarding the Accumulation Points of Sets in a Topological Space

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

Equivalently, for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$.

Furthermore, the set of all accumulation points of $A$ is denoted as $A'$.

We will now look at some basic theorems regarding accumulation points 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 $A' \subseteq B'$.
• Proof: Suppose that $A \subseteq B$ and let $x \in A'$. Then $x$ is an accumulation point of $A$ and so for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$. Since $A \subseteq B$, we see that:
(1)
\begin{align} \quad A \cap U \setminus \{ x \} \subseteq B \cap U \setminus \{ x \} \end{align}
• So for all $U \in \tau$ with $x \in U$ we have that $B \cap U \setminus \{ x \} \neq \emptyset$, so $x$ is an accumulation point of $B$, i.e., $x \in B'$. Therefore $A' \subseteq B'$. $\blacksquare$
 Theorem 2: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $A' \cup B' = (A \cup B)'$.
• Proof: Let $x \in A' \cup B'$. Then $x \in A'$ and $x \in B'$. Since $x \in A'$ we have that $x$ is an accumulation point of $A$ and so for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$. Similarly, since $x \in B'$ we have that $x$ is an accumulation point of $B$ and so for all $U \in \tau$ with $x \in U$ we have that $B \cap U \setminus \{ x \} \neq \emptyset$. Now:
(2)
\begin{align} \quad [A \cap U \setminus \{ x \} ] \cup [B \cap U \setminus \{ x \} ] = (A \cup B) \cap U \setminus \{ x \} \neq \emptyset \end{align}
• So for all $U \in \tau$ with $x \in U$ we have that $(A \cup B) \cap U \setminus \{ x \} \neq \emptyset$ so $x$ is an accumulation point of $A \cup B$, so $x \in (A \cup B)'$. Therefore:
(3)
\begin{align} \quad A' \cup B' \subseteq (A \cup B)' \end{align}
• Now let $x \in (A \cup B)'$ and suppose that $x \not \in A' \cup B'$. Then $x \not \in A'$ and $x \not \in B'$, so $x$ is not an accumulation point of $A$ and $x$ is not an accumulation point of $B$. Since $x$ is not an accumulation point of $A$ there exists a $U_A \in \tau$ with $x \in U_A$ such that $A \cap U_A \setminus \{ x \} = \emptyset$. Similarly, since $x$ is not an accumulation point of $B$ there exists a $U_B \in \tau$ with $x \in U_B$ such that $B \cap U_B \setminus \{ x \} = \emptyset$. Let $U = U_A \cup U_B$. Then $U \in \tau$ since $U_A, U_B \in \tau$ and clearly $x \in U$. So for $x \in U$ we have that:
(4)
\begin{align} \quad U \cap (A \cup B) \setminus \{ x \} = (U_A \cup U_B) \cap (A \cup B) \setminus \{ x \} = (U_A \cap A \setminus \{ x \} ) \cup (U_B \cap B \setminus \{ x \} ) = \emptyset \cup \emptyset = \emptyset \end{align}
• Therefore $x \not \in (A \cup B)'$, a contradiction. Therefore the assumption that $x \not \in A' \cup B'$ was false and so $x \in A' \cup B'$ so:
(5)
\begin{align} \quad (A \cup B)' \subseteq A' \cup B' \end{align}
• We conclude that $A' \cup B' = (A \cup B)'$. $\blacksquare$
 Theorem 3: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $A' \cap B' = (A \cap B)'$.
• Proof: Let $x \in A' \cap B'$. Then $x \in A'$ and $x \in B'$. Since $x \in A'$ we have that $x$ is an accumulation point of $A$ so for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$. Similarly, since $x \in B'$ we have that $x$ is an accumulation point of $B$ so for all $U \in \tau$ with $x \in U$ we have that $B \cap U \setminus \{ x \} \neq \emptyset$. So:
(6)
\begin{align} \quad [A \cap U \setminus \{ x \}] \cap [B \cap U \setminus \{ x \}] = (A \cap B) \cap U \setminus \{ x \} \neq \emptyset \end{align}
• Hence for all $U \in \tau$ with $x \in U$ we have that $(A \cap B) \cap U \setminus \{ x \} \neq \emptyset$ so $x$ is an accumulation point of $A \cap B$ so $x \in (A \cap B)'$. Therefore:
(7)
\begin{align} \quad A' \cap B' \subseteq (A \cap B)' \end{align}
• Now let $x \in (A \cap B)'$. Then for all $U \in \tau$ with $x \in U$ we have that $(A \cap B) \cap U \setminus \{ x \} \neq \emptyset$. So:
(8)
\begin{align} \quad (A \cap B) \cap U \setminus \{ x \} = [A \cap U \setminus \{ x \} ] \cap [B \cap U \setminus \{ x \}] \neq \emptyset \end{align}
• Since the intersection of the two sets above is nonempty, it follows that both sets in the intersection are nonempty. Hence $A \cap U \setminus \{ x \} \neq \emptyset$ and $B \cap U \setminus \{ x \} \neq \emptyset$. Therefore $x$ is an accumulation point of $A$ and $x$ is an accumulation point of $B$, so $x \in A'$ and $x \in B'$, i.e., $x \in A' \cap B'$ and so:
(9)
\begin{align} \quad (A \cap B)' \subseteq A' \cap B' \end{align}
• Hence we conclude that $A' \cap B' = (A \cap B)'$. $\blacksquare$