Bases for a Topology

# Bases for a Topology

 Definition: Let $(X, \tau)$ be a topological space. Let $x \in X$. A Base at $x$ (or Local Base at $x$) is a collection $\mathcal B_x$ of open neighbourhoods of $x$ such that for every open neighbourhood $U$ of $x$ there is a $B \in \mathcal B_x$ for which $x \in B_x \subseteq U$. A Base for the Topology $\tau$ is a collection of open sets $\mathcal B \subseteq \tau$ which contains a base at $x$ for each $x \in X$.

Observe that $\mathcal B = \tau$ is always a base for the topology $\tau$ albeit not that interesting.

For example, consider the set $\mathbb{R}$ with the usual Euclidean topology. There are many types of open sets in $\mathbb{R}$. For each $x \in \mathbb{R}$, an example of a base at $x$ is the collection $\mathcal B_x = \{ B(x, r) : r > 0 \}$ of open balls entered at $x$. Thus, a base for the usual topology on $\mathbb{R}$ is the collection of all open balls of various centers and radii.

In general, if $(X, \| \cdot \|_X)$ is a normed linear space then the topology on $X$ induced by the norm (i.e., the metric topology on $X$ induced by the metric $d(x, y) = \| x - y \|_X$) is such that for each $x \in X$ the collection $\mathcal B_x = \{ B_X(x, r) : r > 0 \}$ of open balls where $B_X(x, r) = \{ y \in X : \| x - y \|_X < r \}$ is a base at $x$, and the collection $\mathcal B$ of all such open balls of various centers and radii is a base for the topology induced by the norm.

 Proposition 1: Let $(X, \tau)$ be a topological space. Then $\mathcal B \subseteq \tau$ is a base for the topology $\tau$ if and only if every open set $U \in \tau$ can be expressed as a union of sets in $\mathcal B$.

By convention, we will say that $\emptyset$ can be expressed as the empty union.

• Proof: $\Rightarrow$ Suppose that $\mathcal B$ is a base for the topology $\tau$. Let $U \in \tau$. For each $x \in U$ there is a base at $x$, call it $\mathcal B_x$. Since $x \in U$ and $U \in \tau$ by definition there exists a $B_x \in \mathcal B$ such that $x \in B_x \subseteq U$. Then $\displaystyle{U = \bigcup_{x \in X} B_x}$ and each $B_x \in \mathcal B_X \subseteq \mathcal B$.
• $\Leftarrow$ Suppose that every open set can be expressed as a union of sets in $\mathcal B$. If $x \in X$ and $U \in \tau$ is such that $x \in U$ then since $U$ is open it can be expressed as a union of sets from $\mathcal B$. For each $U$ select a nonempty set in the expression of $U$ as a union of sets from $\mathcal B$. Let $\mathcal B_x$ be the collection of such sets. Then $\mathcal B_x$ is a base at $x$ since for each $U \in \tau$ with $x \in U$ there is a $B \in \mathcal B_x$ such that $x \in B \subseteq U$. Furthermore, $\mathcal B_x \subseteq \mathcal B$. So $\mathcal B$ is a base for the topology $\tau$. $\blacksquare$

From proposition 1 above, if a base $\mathcal B$ for a topology $\tau$ is given then the topology $\tau$ is completely defined. The topology will consist of $\emptyset$ and all unions of sets from $\mathcal B$. For this reason, it is extremely convenient to describe a topology by a base. The next question to ask is whether a given collection of sets is a base for some topology on $X$. The following proposition tells us when that’s possible.

 Proposition 2: Let $X$ be a nonempty set and let $\mathcal B$ be a collection of subsets of $X$. Then $\mathcal B$ is a base for some topology $\tau$ on $X$ if and only if the following conditions hold: 1) $\displaystyle{X = \bigcup_{B \in \mathcal B} B}$. 2) For all $B_1, B_2 \in \mathcal B$ and all $x \in B_1 \cap B_2$ there exists a $B \in \mathcal B$ for which $x \in B \subseteq B_1 \cap B_2$.