IFF Criterion for Net Convergence Uniqueness

# IFF Criterion for Net Convergence Uniqueness

Theorem 1: Let $(X, \tau)$ be a topological space. Then every net in $X$ which net converges will net converge to a unique point in $X$ if and only if $X$ is a Hausdorff space. |

**Proof:**$\Rightarrow$ (Contrapositive) Suppose that $X$ is not a Hausdorff space. Then there exists a pair of points $s, t \in X$ with $s \neq t$ such that every open neighbourhood $U$ of $s$ and every open neighbourhood $V$ of $t$ is such that $U \cap V \neq \emptyset$.

- Out goal is to construct a net $(S, \leq) = \{ S_n : n \in D, \leq \}$ that net converges to both $s$ and $t$ in order to prove the contrapositive.

- Let $\mathcal F_s$ be the collection of all open neighbourhoods of $s$ and let $\mathcal F_t$ be the collection of all open neighbourhoods of $t$. Let $\subseteq$ denote the inclusion relation. Then $(\mathcal F_s, \subseteq)$ and $(\mathcal F_t, \subseteq)$ are both directed sets.

- Consider the set:

\begin{align} \quad \mathcal F_s \times \mathcal F_t = \{ (U, V) : U \in \mathcal F_s, V \in \mathcal F_t \} \end{align}

- Define a relation $\leq$ on this set for all $(A, B), (C, D) \in \mathcal F_s \times \mathcal F_t$ by:

\begin{align} \quad (A, B) \leq (C, D) \quad \mathrm{if \: and \: only \: if} \quad A \supseteq C \: \mathrm{and} \: B \supseteq D \end{align}

- It is easy to check that then $(\mathcal F_s \times \mathcal F_t, \leq)$ is a directed set. Indeed, $(A, B) \leq (A, B)$ for all $(A, B) \in \mathcal F_s \times \mathcal F_t$ since $A \supseteq A$ and $B \supseteq B$ (and so reflexivity is satisfied). Also, if $(A, B) \leq (C, D)$ and $(C, D) \leq (E, F)$ then $A \supseteq C \supseteq E$ and $B \supseteq D \supseteq F$, and so $(A, B) \leq (E, F)$ (that is, transitivity is satisfied). Lastly, if $(A, B), (C, D) \in \mathcal F_s \times \mathcal F_t$ then $A \cap C$ is an open neighbourhood of $s$, and $B \cap D$ is an open neighbourhood of $t$, and $(A, B) \leq (A \cap C, B \cap D)$ and $(C, D) \leq (A \cap C, B \cap D)$, so $\mathcal F_s \times \mathcal F_t$ possesses an upper bound for of its pairs of elements.

- Now observe that if $(U, V) \in \mathcal F_s \times \mathcal F_t$ then $U \cap V \neq \emptyset$ by assumption. Take any point $S_{(U, V)} \in U \cap V$.

- Let $(S, \leq) = \{ S_n : n = (V, W) \in \mathcal F_s \times \mathcal F_t, \leq \}$ be a net in $X$ defined this way. Then for any open neighbourhood $U$ of $s$ and any open neighbourhood $V$ of $t$ we have that if $n = (V, W)$ is such that $(U, V) \leq (W, Y)$ then $S_{(W, Y)} \in W \cap Y \subseteq U \cap V$.

- But $U \cap V \subseteq U$ and $U \cap V \subseteq V$. Therefore $(S, \leq)$ is eventually in both $U$ and $V$. Since this holds true for all open neighbourhoods $U$ of $s$ and $V$ of $t$ we have that $(S, \leq)$ net converges to both $s$ and $t$, which are distinct points. This proves the contrapositive.

- $\Leftarrow$ Suppose that $X$ is a Hausdorff space. Let $s, t \in X$ be such that $s \neq t$. Let $(S, \leq) = \{ S_n : n \in D, \leq \}$ be net in $X$ that net converges to $s$ and that net converges to $t$.

- Since $X$ is a Hausdorff space and $s \neq t$ there exists open neighbourhoods $U$ of $s$ and $V$ of $t$ such that $U \cap V = \emptyset$. Now since $(S, \leq)$ net converges to $s$ there exists an $N_1 \in D$ such that if $N_1 \leq n$ then $S_n \in U$, and since $(S, \leq)$ net converges to $t$ there exists an $N_2 \in D$ such that if $N_2 \leq n$ then $S_n \in V$.

- Let $N = \max \{ N_1, N_2 \}$. Then if $N \leq n$ we have that $S_n \in U$ and $S_n \in V$. That is, $S_n \in U \cap V$. But $U \cap V = \emptyset$ - a contradiction. Therefore the assumption that $(S, \leq)$ net converges to two distinct points is false. $\blacksquare$

Since points of net convergence are unique in Hausdorff topological spaces (whenever a net, net-converges), we introduce the following notation.

Definition: Let $(X, \tau)$ be a Hausdorff topological space. If $(S, \leq) = \{ S_n : n \in D, \leq \}$ is a net that net converges to $s \in X$ then we say that $s$ is the Limit of the net and we write $\displaystyle{\tau-\lim \{ S_n : n \in D, \leq \} = s}$. |