Proof of Basis Topology Generation

We provide a detailed, self-contained proof that a collection of sets satisfying the basis conditions generates a unique topology. This proof constructs the topology explicitly and verifies all the axioms, making it a model for similar verification arguments throughout topology.

Setup and Statement

Theorem1.6Topology Generated by a Basis

Let $X$ be a set and $\mathcal{B}$ a collection of subsets of $X$ satisfying:

(B1) For each $x \in X$ , there exists $B \in \mathcal{B}$ with $x \in B$ .
(B2) For each $B_1, B_2 \in \mathcal{B}$ and each $x \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$ .

Then the collection $\tau_{\mathcal{B}} = \left\{U \subseteq X : \forall x \in U, \exists B \in \mathcal{B}, \; x \in B \subseteq U\right\}$ is a topology on $X$ , and $\mathcal{B}$ is a basis for $\tau_{\mathcal{B}}$ . Moreover, $\tau_{\mathcal{B}}$ consists precisely of all unions of subcollections of $\mathcal{B}$ .

Complete Proof

Proof

We verify each topology axiom in detail.

Step 1: $\emptyset \in \tau_{\mathcal{B}}$ .

The condition " $\forall x \in U, \exists B \in \mathcal{B}, x \in B \subseteq U$ " is vacuously true when $U = \emptyset$ (there is no $x \in \emptyset$ ). Hence $\emptyset \in \tau_{\mathcal{B}}$ .

Step 2: $X \in \tau_{\mathcal{B}}$ .

Let $x \in X$ . By (B1), there exists $B \in \mathcal{B}$ with $x \in B$ . Since $B \subseteq X$ , we have $x \in B \subseteq X$ . As $x$ was arbitrary, $X \in \tau_{\mathcal{B}}$ .

Step 3: $\tau_{\mathcal{B}}$ is closed under arbitrary unions.

Let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary family with each $U_\alpha \in \tau_{\mathcal{B}}$ . Set $U = \bigcup_{\alpha \in A} U_\alpha$ . Let $x \in U$ . Then $x \in U_\alpha$ for some $\alpha \in A$ . Since $U_\alpha \in \tau_{\mathcal{B}}$ , there exists $B \in \mathcal{B}$ with $x \in B \subseteq U_\alpha \subseteq U$ . Thus $U \in \tau_{\mathcal{B}}$ .

Step 4: $\tau_{\mathcal{B}}$ is closed under finite intersections.

It suffices to show closure under binary intersections; the general case follows by induction (the base case $n = 0$ gives the empty intersection $X \in \tau_{\mathcal{B}}$ ).

Let $U_1, U_2 \in \tau_{\mathcal{B}}$ and let $x \in U_1 \cap U_2$ . Since $U_1 \in \tau_{\mathcal{B}}$ , there exists $B_1 \in \mathcal{B}$ with $x \in B_1 \subseteq U_1$ . Since $U_2 \in \tau_{\mathcal{B}}$ , there exists $B_2 \in \mathcal{B}$ with $x \in B_2 \subseteq U_2$ . Now $x \in B_1 \cap B_2$ , and by (B2), 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 $U_1 \cap U_2 \in \tau_{\mathcal{B}}$ .

Step 5: $\mathcal{B}$ is a basis for $\tau_{\mathcal{B}}$ .

First, every $B \in \mathcal{B}$ belongs to $\tau_{\mathcal{B}}$ : for $x \in B$ , we have $x \in B \subseteq B$ , so the defining condition is satisfied with $B$ itself. Thus $\mathcal{B} \subseteq \tau_{\mathcal{B}}$ .

Next, for every $U \in \tau_{\mathcal{B}}$ and every $x \in U$ , by definition there exists $B \in \mathcal{B}$ with $x \in B \subseteq U$ . This is precisely the condition that $\mathcal{B}$ is a basis for $\tau_{\mathcal{B}}$ .

Step 6: $\tau_{\mathcal{B}}$ equals the collection of all unions of elements of $\mathcal{B}$ .

Let $\mathcal{U} = \left\{\bigcup_{\beta \in B} B_\beta : B_\beta \in \mathcal{B}\right\}$ denote the set of all unions of subcollections of $\mathcal{B}$ .

$(\mathcal{U} \subseteq \tau_{\mathcal{B}})$ : Each $B_\beta \in \mathcal{B} \subseteq \tau_{\mathcal{B}}$ , and $\tau_{\mathcal{B}}$ is closed under unions, so $\bigcup B_\beta \in \tau_{\mathcal{B}}$ .

$(\tau_{\mathcal{B}} \subseteq \mathcal{U})$ : Let $U \in \tau_{\mathcal{B}}$ . For each $x \in U$ , choose $B_x \in \mathcal{B}$ with $x \in B_x \subseteq U$ . Then $U = \bigcup_{x \in U} B_x$ , which is a union of elements of $\mathcal{B}$ . So $U \in \mathcal{U}$ .

Step 7: Uniqueness.

Suppose $\tau$ is any topology on $X$ having $\mathcal{B}$ as a basis. Then $U \in \tau$ if and only if for every $x \in U$ , there exists $B \in \mathcal{B}$ with $x \in B \subseteq U$ , which is precisely the defining condition for $\tau_{\mathcal{B}}$ . Thus $\tau = \tau_{\mathcal{B}}$ .

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Illustrative Verification

ExampleVerifying the Metric Topology Basis

Let $(X, d)$ be a metric space. The collection of open balls $\mathcal{B} = \{B_d(x, r) : x \in X, r > 0\}$ satisfies the basis conditions:

(B1): For any $x \in X$ , we have $x \in B_d(x, 1)$ .

(B2): Let $y \in B_d(x_1, r_1) \cap B_d(x_2, r_2)$ . Set $\epsilon = \min(r_1 - d(y, x_1), r_2 - d(y, x_2)) > 0$ . Then $B_d(y, \epsilon) \subseteq B_d(x_1, r_1) \cap B_d(x_2, r_2)$ , since for $z \in B_d(y, \epsilon)$ : $d(z, x_i) \leq d(z, y) + d(y, x_i) < \epsilon + d(y, x_i) \leq r_i.$

Thus, $\mathcal{B}$ generates a topology on $X$ , the metric topology.

ExampleThe K-Topology on the Reals

Let $K = \{1/n : n \in \mathbb{Z}^+\}$ . The collection $\mathcal{B}_K = \{(a, b) : a < b\} \cup \{(a, b) \setminus K : a < b\}$ is a basis for a topology on $\mathbb{R}$ called the $K$ -topology.

To verify (B2), one checks all cases of intersections. The most interesting case is when $x \in (a_1, b_1) \cap ((a_2, b_2) \setminus K)$ . Then $x \notin K$ , and the intersection $(a_1, b_1) \cap ((a_2, b_2) \setminus K) = ((\max(a_1, a_2), \min(b_1, b_2)) \setminus K)$ , which is in $\mathcal{B}_K$ .

The $K$ -topology is strictly finer than the standard topology. It provides a Hausdorff space that is not regular, demonstrating that separation axioms form a strict hierarchy.

RemarkConstructive Nature of the Proof

This proof is fully constructive: given a specific point and specific open sets, each step provides an explicit basis element. This constructive character makes the basis criterion especially useful in practice. When defining a new topology, one specifies a basis, verifies (B1) and (B2), and immediately obtains a well-defined topological space without needing to enumerate all open sets.