Net Criterion for a Point to be an Accumulation Point of a Set

# Net Criterion for a Point to be an Accumulation Point of a Set

Recall from the Convergence of Sequences and Nets in Topological Spaces that if $(X, \tau)$ is a topological space then a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ is said to converge to $s \in X$ if for every open neighbourhood $B$ of $s$, $(S, \leq)$ is eventually in $B$. That is, for every open neighbourhood $B$ of $s$ there exists an $N \in D$ such that if $N \leq n$ then $S_n \in B$.

We will now prove a criterion for a point to be an accumulation point of a set.

Theorem 1 (Net Convergence Criterion for Accumulation Points of a Set): Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $a \in X$ is an accumulation point of $A$ if and only if there exists a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ in $A \setminus \{ a \}$ that net converges to $a$. |

**Proof:**$\Rightarrow$ Let $a$ be an accumulation point of $A$. Then every open neighbourhood $B$ of $a$ contains a point in $A$ different from $a$. That is:

\begin{align} \quad B \cap (A \setminus \{ a \}) \neq \emptyset \end{align}

- For for any open neighbourhood $B$ of $a$ let $a_B \in B \cap (A \setminus \{ a \})$.

- Let $\mathcal F$ be the collection of all open neighbourhoods of $a$. Then $(\mathcal F, \subseteq)$ is a directed set. Let $S$ be the function defined on $D = \mathcal F$ defined for each $B \in D$ by:

\begin{align} \quad S_B = a_B \end{align}

- Then $(S, \subseteq) = \{ S_n : n \in D, \subseteq \}$ is a net in $A \setminus \{ a \}$. Let $U$ be any open neighbourhood of $a$. Then observe that if $U \subseteq V$ then $a_U \in U \cap (A \setminus \{ a \})$ is such that $a_U \in V \cap (A \setminus \{ a \})$. In otherwords, if $U \subseteq V$ then $S_V \in U$. Therefore $(S, \subseteq)$ net converges to $a$. So there exists a net in $A \setminus \{ a \}$ that net converges to $a$.

- $\Leftarrow$ Suppose that there exists a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ in $A \setminus \{ a \}$ that net converges to $a$. Then for every open neighbourhood $B$ of $a$, the net $(S, \leq)$ is eventually in $A$. In other words, there exists an $N \in D$ such that if $N \leq n$ then $S_n \in A \setminus \{ a \}$. But then for every open neighbourhood $B$ of $a$ we have that:

\begin{align} \quad B \cap (A \setminus \{ a \}) \neq \emptyset \end{align}

- So $a$ is an accumulation point of $A$. $\blacksquare$

Note: When dealing with $\mathbb{R}$ with the usual topology, recall that if $A \subseteq \mathbb{R}$ then a point $a \in \mathbb{R}$ is an accumulation point of $A$ if and only if there exists a sequence in $A \setminus \{ a \}$ that converges to $a$. Compare and contrast with the theorem above which extends to any general topological space. |