Subsequences and Subnets in a Topological Space

# Subsequences and Subnets in a Topological Space

## Subsequences

Definition: Let $(X, \tau)$ be a topological space and let $(x_n)$ be a sequence in $X$. Another sequence $(y_n)$ in $X$ is said to be a Subsequence of $(x_n)$ if there exists a function $f : \mathbb{N} \to \mathbb{N}$ such that $y_n = x_{f(n)}$ for every $n \in \mathbb{N}$ and $f$ is an increasing function (not necessarily strictly increasing). |

## Subnets

Definition: Let $(X, \tau) $] be a topological space and let [[$ (S, \leq) = \{ S_n : n \in D, \leq \}$ be a net in $X$. Another net $(T, \leq) = \{ T_n : n \in E, \leq \}$ (with the same order relation $\leq$) is said to be a Subnet of $(S, \leq)$ if there exists a function $f : E \to D$ such that $T_n = S_{f(n)}$ for all $n \in E$ and such that for every $n \in D$ there exists an $m \in E$ such that whenever $m \leq p$ we have that $n \leq f(p)$. |

*The second condition in the definition of a subnet above simply says that if $p$ gets large (with respect to the relation $\leq$, then $f(p)$ gets large.*

Proposition 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then if $(S, \leq)$ is a net that is eventually in $A$ and $(T, \leq)$ is a subset of $(S, \leq)$ then $(T, \leq)$ is eventually in $A$. |

**Proof:**Suppose that $(S, \leq) = \{ S_n : n \in D, \leq \}$ is eventually in $A$. Let $(T, \leq) = \{ T_n : n \in E, \leq \}$ be a subnet of $(S, \leq)$. Then there exists a function $f : E \to D$ such that $T_n = S_{f(n)}$ for all $n \in E$ and is such that for every $n \in D$ there exists an $m \in E$ for which $m \leq p$ implies $n \leq f(p)$.

- Since $(S, \leq)$ is eventually in $A$ there exists an $n \in D$ such that if $N \leq m$ then $S_n \in A$. Let $M$ be such that if $M \leq p$ then $N \leq f(p)$. Then if $M \leq p$ we have that

\begin{align} \quad T_p = S_{f(p)} \in A \end{align}

- So $(T, \leq)$ is eventually in $A$. $\blacksquare$