Bases of Neighbourhoods for a Point

# Bases of Neighbourhoods for a Point

Definition: Let $(E, \tau)$ be a topological space and let $x \in E$. A Base of Neighbourhoods for $x$ is a collection $\mathcal V_x$ of neighbourhoods of $x$ such that for each $U \in \mathcal U_x$, there exists a $V \in \mathcal V_x$ with $x \in V \subseteq U$. |

*Recall that $\mathcal U_x$ is defined to be the collection of all neighbourhoods of $x$.*

For example, consider the space $\mathbb{R}$ and let $0 \in \mathbb{R}$. Then $\mathcal U_x$ consists of many sets. For example, $[-1, 1]$, $(-1, 2)$, $(-\infty, 2) \cup \{3 \}$, and so forth, are all neighbourhoods of $x$. It is easy to see that the collection $\mathcal V_x := \left \{ \left ( -\frac{1}{n}, \frac{1}{n} \right ) : n \in \mathbb{N} \right \}$ is a base of neighbourhoods for $0$. The collection $\mathcal W_x := \left \{ \left [ - \frac{1}{n}, \frac{1}{n} \right ] : n \in \mathbb{N} \right \}$ is also a base of neighbourhoods for $0$.

Proposition 1: Let $(E, \tau)$ be a topological space and let $x \in E$. Then, the collection of all open neighbourhoods of $x$ is a base of neighbourhoods for $x$. |

**Proof:**Let $\mathcal V_x$ be the collection of all open neighbourhoods of $x$. Given a neighbourhood $U \in \mathcal U_x$ of $x$, by definition, there exists an open set $V$ such that $x \in V \subseteq U$. But then $V \in \mathcal V_x$. So indeed, $\mathcal V_x$ is a base of neighbourhoods for $x$. $\blacksquare$

## First Countable Topological Spaces

Definition: A topological space $(E, \tau)$ is said to be First Countable if for every $x \in E$, $x$ has a countable base of neighbourhoods. |