Net Criterion for a Point to be in the Closure of a Set

Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $a \in \mathrm{A}$ if and only if there exists a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ in $A$ that net converges to $a$.

• Proof: $\Rightarrow$ Suppose that $a \in \overline{A}$. Recall that $\overline{A} = A \cup A'$.

• If $a \in A$ then let $(S, \leq) = \{ S_n : n \in D, \leq \}$ be defined for all $n \in D$ by $S_n = a$. Then $(S, \leq)$ net converges to $a$ since every open neighbourhood of $a$ clearly contains $a$ by definition (and thus, $(S, \leq)$ is eventually in every open neighbourhood of $a$).

• If $a \in A'$ then by the Net Criterion for a Point to be an Accumulation Point of a Set we have that there exists a net $(S, \leq)$ in $A \setminus \{ a \} \subseteq A$ that net converges to $a$.

• $\Leftarrow$ Suppose that there exists a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ in $A$ that net converges to $a$. Then for every open neighbourhood $U$ of $a$ there exists an $N \in D$ such that if $N \leq n$ then $S_n \in U$. In particular, for every open neighbourhood $U$ of $a$ we have that:

(1)

\begin{align} \quad U \cap A \neq \emptyset \end{align}

• But this is equivalent to saying that $a \in \bar{A}$. $\blacksquare$

Corollary 2: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is closed if and only if whenever a net $(S, \leq)$ net converges, it net converges to a point in $A$.