Basic Thms. Regarding Local Bases of a Point in a Topological Space

# Basic Theorems Regarding Local Bases of a Point in a Topological Space

Recall from the Local Bases of a Point in a Topological Space page that if $(X, \tau)$ is a topological space the a local basis for a point $x \in X$ is a collection of open neighbourhoods of $x$ denoted $\mathcal B_x$ such that for all open neighbourhoods of $x$, $U \in \tau$ with $x \in U$ there exists a $B \in \mathcal B_x$ such that:

(1)\begin{align} \quad x \in B \subseteq \mathcal B_x \end{align}

We will now look at some basic theorems regarding local bases of a point in a topological space.

Theorem 1: Let $(X, \tau)$ be a topological space and let $\mathcal B$ be a basis of $\tau$. Then for each $x \in X$, $\mathcal B_x = \{ B \in \mathcal B : x \in B \}$ is a local basis of $x$. |

**Proof:**Let $x \in X$ and let $\mathcal B_x = \{ B \in \mathcal B : x \in B \}$. To show that $\mathcal B_x$ is a local basis of $x$ we need to show that for all $U \in \tau$ with $x \in U$ that there exists a $B \in \mathcal B_x$ such that $x \in B \subseteq U$.

- Let $U \in \tau$ with $x \in U$. Since $\mathcal B$ is a basis of $\tau$ we have that there exists a subcollection $\mathcal B^* \subseteq \mathcal B$ such that:

\begin{align} \quad U = \bigcup_{B \in \mathcal B^*} B \end{align}

- Since $x \in U$ we have that $x \in \bigcup_{B \in \mathcal B^*} B$. So $x$ is contained in at least one $B \in \mathcal B^*$. Let $B_x \in \mathcal B^*$ be such that $x \in B_x$. Then $B_x \in \mathcal B_x$ since $x \in B_x$ and $B_x \in \mathcal B^* \subseteq \mathcal B$ and furthermore:

\begin{align} \quad x \in B_x \subseteq \bigcup_{B \in \mathcal B^*} B = U \end{align}

- So, for all $U \in \tau$ with $x \in U$ there exists a $B_x \in \mathcal B_x = \{ B \in \mathcal B : x \in B \}$ such that $x \in B_x \subseteq U$. Therefore, $\mathcal B_x$ is a local basis of $x$. $\blacksquare$

Theorem 2: Let $(X, \tau)$ be a topological space. If $\{ \mathcal B_x \}_{x \in X}$ is a collection of local bases for each $x \in X$ then $\mathcal B = \bigcup_{x \in X} \mathcal B_x$ is a basis for the topology $\tau$. |

**Proof:**Let $\mathcal B = \bigcup_{B \in \mathcal B} B$. Recall that a set $\mathcal B$ is a basis for the topology $\tau$ if every $U \in \tau$ is the union of some subcollection of sets from $\mathcal B$, i.e., for every $U \in \tau$ there exists a $\mathcal B^* \subseteq \mathcal B$ such that $U = \bigcup_{B \in \mathcal B} B$.

- Let $U \in \tau$ and let $x \in U$. Then there exists a local basis of $x$, $\mathcal B_x \subseteq \mathcal B$. Since $\mathcal B_x$ is a local basis of $x$, we have that for $U \in \tau$ there exists a $B_x \in \mathcal B_x$ such that:

\begin{align} \quad x \in B_x \subseteq U \end{align}

- So we have that:

\begin{align} \quad U = \bigcup_{x \in U} B_x \end{align}

- Hence all $U \in \tau$ can be expressed as the union of elements from $\mathcal B$, so $\mathcal B = \bigcup_{x \in X} \mathcal B_x$ is a basis for the topology $\tau$. $\blacksquare$