Basis Criterion for Topology

The basis criterion provides a precise characterization of when a collection of subsets constitutes a basis for a topology, and relates different topologies through their bases. This theorem is fundamental for verifying that constructions yield valid topological spaces.

Statement

Theorem1.1Basis Criterion

Let $X$ be a set and let $\mathcal{B}$ be a collection of subsets of $X$ . Then $\mathcal{B}$ is a basis for some topology on $X$ if and only if:

$\mathcal{B}$ covers $X$ : for every $x \in X$ , there exists $B \in \mathcal{B}$ such that $x \in B$ .
For every $B_1, B_2 \in \mathcal{B}$ and every $x \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$ .

Moreover, when these conditions hold, the collection $\tau_{\mathcal{B}} = \left\{ U \subseteq X : \forall x \in U, \; \exists B \in \mathcal{B} \text{ with } x \in B \subseteq U \right\}$ is the unique topology on $X$ for which $\mathcal{B}$ is a basis, and $\tau_{\mathcal{B}}$ equals the set of all unions of elements of $\mathcal{B}$ .

Proof

(Necessity) Suppose $\mathcal{B}$ is a basis for a topology $\tau$ on $X$ . Since $X \in \tau$ , and $X$ is a union of basis elements, condition (1) holds. For condition (2), if $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$ , then $B_1 \cap B_2 \in \tau$ (as $B_1, B_2$ are open and $\tau$ is closed under finite intersections). Since $\mathcal{B}$ is a basis, there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$ .

(Sufficiency) Assume conditions (1) and (2). Define: $\tau_{\mathcal{B}} = \left\{ U \subseteq X : \forall x \in U, \; \exists B \in \mathcal{B} \text{ with } x \in B \subseteq U \right\}.$

We verify the topology axioms:

Empty set and whole space: $\emptyset \in \tau_{\mathcal{B}}$ vacuously. By condition (1), $X \in \tau_{\mathcal{B}}$ .

Arbitrary unions: Let $\{U_\alpha\}_{\alpha \in A} \subseteq \tau_{\mathcal{B}}$ and $U = \bigcup_\alpha U_\alpha$ . For $x \in U$ , there exists $\alpha$ with $x \in U_\alpha$ . Since $U_\alpha \in \tau_{\mathcal{B}}$ , there exists $B \in \mathcal{B}$ with $x \in B \subseteq U_\alpha \subseteq U$ . Thus $U \in \tau_{\mathcal{B}}$ .

Finite intersections: It suffices to check binary intersections (by induction). Let $U_1, U_2 \in \tau_{\mathcal{B}}$ and $x \in U_1 \cap U_2$ . There exist $B_1, B_2 \in \mathcal{B}$ with $x \in B_1 \subseteq U_1$ and $x \in B_2 \subseteq U_2$ . By condition (2), there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2 \subseteq U_1 \cap U_2$ . Thus $U_1 \cap U_2 \in \tau_{\mathcal{B}}$ .

Equivalence of descriptions: Every $U \in \tau_{\mathcal{B}}$ satisfies $U = \bigcup_{x \in U} B_x$ where $B_x \in \mathcal{B}$ with $x \in B_x \subseteq U$ . Conversely, any union of elements of $\mathcal{B}$ belongs to $\tau_{\mathcal{B}}$ .

Uniqueness: If $\tau$ is any topology for which $\mathcal{B}$ is a basis, then $U \in \tau$ if and only if $U$ is a union of elements of $\mathcal{B}$ , which equals $\tau_{\mathcal{B}}$ .

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Comparing Topologies via Bases

Theorem1.2Comparison of Topologies via Bases

Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be bases for topologies $\tau_1$ and $\tau_2$ on $X$ , respectively. Then $\tau_1 \subseteq \tau_2$ if and only if for every $B_1 \in \mathcal{B}_1$ and every $x \in B_1$ , there exists $B_2 \in \mathcal{B}_2$ such that $x \in B_2 \subseteq B_1$ .

Proof

( $\Rightarrow$ ) If $\tau_1 \subseteq \tau_2$ , then every $B_1 \in \mathcal{B}_1 \subseteq \tau_1 \subseteq \tau_2$ . Since $\mathcal{B}_2$ is a basis for $\tau_2$ , for every $x \in B_1$ there exists $B_2 \in \mathcal{B}_2$ with $x \in B_2 \subseteq B_1$ .

( $\Leftarrow$ ) Let $U \in \tau_1$ . For any $x \in U$ , there exists $B_1 \in \mathcal{B}_1$ with $x \in B_1 \subseteq U$ , and then by hypothesis there exists $B_2 \in \mathcal{B}_2$ with $x \in B_2 \subseteq B_1 \subseteq U$ . Thus $U \in \tau_2$ .

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Applications

ExampleStandard vs. Lower Limit Topology

Let $\mathcal{B}_1 = \{(a, b) : a < b\}$ (basis for the standard topology on $\mathbb{R}$ ) and $\mathcal{B}_2 = \{[a, b) : a < b\}$ (basis for the lower limit topology on $\mathbb{R}$ ).

For any $(a, b)$ and $x \in (a, b)$ , we have $x \in [x, b) \subseteq (a, b)$ , so $\mathcal{B}_2$ refines $\mathcal{B}_1$ , giving $\tau_{\text{std}} \subseteq \tau_\ell$ .

However, $[0, 1) \in \mathcal{B}_2$ and $0 \in [0, 1)$ , but there is no open interval $(a, b) \in \mathcal{B}_1$ with $0 \in (a, b) \subseteq [0, 1)$ (any such interval would contain negative numbers). So $\tau_\ell \not\subseteq \tau_{\text{std}}$ , and the lower limit topology is strictly finer.

RemarkPractical Importance

The basis criterion is the standard tool for verifying that a given collection generates a topology. It is used repeatedly in constructions such as:

The product topology (basis of finite intersections of preimages).
The metric topology (basis of open balls).
The order topology (basis of open intervals and rays).
Quotient and identification topologies.