2.2Basis for a Topology

Updated 14 Jun 2026Chapter (PDF)

[2.2.1]Definition(Basis, Basis Elements)#

If $X$ is a set, then a basis for some topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) that satisfy the following conditions.

Covering. For each $x \in X$ , there is some $B \in \mathcal{B}$ such that $x \in B$ . That is, $X \subseteq \bigcup_{B \in \mathcal{B}} B$ .
Refinement. If there exist $B_1, B_2 \in \mathcal{B}$ such that $x \in B_1 \cap B_2$ , there is some $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$ .

[2.2.2]Definition(Topology Generated by Basis)#

If there exists some basis $\mathcal{B}$ for $X$ , we say the topology $\Tau$ generated by $\mathcal{B}$ is as follows:

“A subset $U \subseteq X$ is said to be open (i.e., in $\Tau$ ) if, for each $x \in U$ , there is some $B \in \mathcal{B}$ such that $x \in B \subseteq U$ ”.

[2.2.3]Example(Basis of Open Balls/Boxes)#

Let $\mathcal{B}$ be the collection of all open balls in $\bR^n$ . Of course, each $x \in \bR^n$ is contained in some open ball $B \in \mathcal{B}$ . Now consider some $x \in \bR^n$ contained in two open balls $B_1, B_2 \in \mathcal{B}$ —that is, $x \in B_1 \cap B_2$ . Then we can construct a smaller open ball about $x$ , say $B_3$ , such that $x \in B_3 \subseteq B_1 \cap B_2$ (with $B_3 \in \mathcal{B}$ ), of course.

We can do an analogous procedure, but instead defining $\mathcal{B}$ to be the collection of all open boxes in $\bR^n$ . But note that intersections of boxes are, well, boxes, so we do not need to find a basis element inside the intersection of two other basis elements as done with the basis of open balls.

We provide a schematic for these bases in $\bR^2$ .

[2.2.4]Example(Basis that Generates Discrete Topology)#

If $X$ is any set, then we can define a basis $\mathcal{B}$ comprising every singleton subset of $X$ . Let $U \subseteq X$ be any arbitrary subset of $X$ . Then note, for each $x \in U$ , it's true that $\{x\} \in \mathcal{B}$ , such that $x \in \{x\} \subseteq U$ . Thus all subsets of $X$ are open, meaning the topology generated by $\mathcal{B}$ is in fact the discrete one.

We now verify that a topology generated by a basis is, indeed, a topology.

[2.2.5]Theorem(Basis Generates a Topology)#

Suppose $\mathcal{B}$ is a basis for some set $X$ . Let $\Tau$ be a collection of subsets such that, if some subset $U \subseteq X$ is in $\Tau$ , then for each $x \in U$ , there is some $B \in \mathcal{B}$ such that $x \in B \subseteq U$ . Then $(X, \Tau)$ forms a topological space.

Proof.

First, note $\emptyset \in \Tau$ vacuously. Next, note that for each $x \in X$ , there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq X$ (both by definition of a basis element), so $X \in \Tau$ .

We show arbitrary union is closed in $\Tau$ . Suppose $\{U_\lambda\}_{\lambda \in \Lambda}$ is an indexed family of elements of $\Tau$ . Define

U = \bigcup_{\lambda \in \Lambda} U_\lambda.

If $x \in U$ , then there is some $\lambda \in \Lambda$ such that $x \in U_\lambda$ . Since $U_\lambda$ is open, there exists some $B_\lambda \in \mathcal{B}$ such that $x \in B_\lambda \subseteq U_\lambda \subseteq U$ . Thus $U$ is open.

Finally, we show finite intersection is closed in $\Tau$ . That is, for some $n \in \bZ_+$ , the intersection of open sets $U_1 \cap \cdots U_n$ is open. We prove this inductively, using $n = 2$ as our base case. Suppose $U_1, U_2 \in \Tau$ —we show $U_1 \cap U_2 \in \Tau$ as well. First, suppose $x \in U_1 \cap U_2$ . Then there exist $B_1, B_2 \in \Tau$ such that $x \in B_1 \subseteq U_1$ and $x \in B_2 \subseteq U_2$ . Since $x \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2 \subseteq U_1 \cap U_2$ . Thus $U_1 \cap U_2 \in \Tau$ .

Now suppose this result is true for $n - 1$ . Then note

U_1 \cap \cdots \cap U_n = \underbrace{(U_1 \cap \cdots U_{n - 1})}_{:= U_1} \cap \underbrace{U_n}_{:= U_2}.

$U_1$ is open by hypothesis, of course. Since we have reduced this to $n = 2$ , it follows that the intersection is, indeed, finite. Thus $\Tau$ is a topology!

[2.2.6]Corollary(Intuition Behind Basis)#

Let $X$ be a set, and let $\mathcal{B}$ be a basis for some topology $\Tau$ on $X$ . Then $\Tau$ is the collection of all unions of elements of $\mathcal{B}$ .

Proof.

Let $\{B_\lambda\}_{\lambda \in \Lambda}$ be a collection of elements of $\mathcal{B}$ . Of course, each basis element is an element of the topology it generates. Since topologies are closed under arbitrary union, the union of elements in this collection is a subset of $\Tau$ .

Conversely, suppose $T \in \Tau$ . For each $x \in T$ , there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq T$ . Thus $T \subseteq \bigcup_{x \in T} B_x$ , a union of a collection of elements from $\mathcal{B}$ , completing the proof.

Immediately, after choosing a basis for a topology, we note that every element of the topology may be generated by a union of some elements of the basis. This is reminiscent of vectors being generated by a linear combination of some elements of the basis. But note that each vector is uniquely expressible in a vector space's basis, while each element of a topology does not have a necessarily unique expression in the topology's basis.

[2.2.7]Theorem#

Suppose $(X, \Tau)$ is a topological space. Let $\mathcal{C}$ be a collection of open sets such that, for each open set $U \subseteq X$ (under the topology $\Tau$ ) and for each $x \in U$ , there is some $C \in \mathcal{C}$ such that $x \in \mathcal{C} \subseteq U$ . Then $\mathcal{C}$ is a basis for the topology $\Tau$ .

Proof.

First, we show $\mathcal{C}$ is a basis. Note $X$ is open, so by hypothesis, every $x \in X$ is met with some $C \in \mathcal{C}$ such that $x \in C$ , and of course $C \subseteq X$ . Moreover, suppose $x \in X$ is such that, for some $C_1, C_2 \in \mathcal{C}$ , it follows that $x \in C_1 \cap C_2$ . Since $C_1, C_2$ are open under $\Tau$ , it follows that $C_1 \cap C_2$ are open under $\Tau$ , so there exists some $C_3 \in \mathcal{C}$ such that $x \in C_3 \subseteq C_1 \cap C_2$ by hypothesis. Thus $\mathcal{C}$ is a basis.

Now we show the topology generated by $\mathcal{C}$ , say $\Tau'$ , is $\Tau$ itself. This is slightly tautological, but whatever. Suppose $U \subseteq X$ is open under $\Tau$ . The, for each $x \in U$ , there is some $C \in \mathcal{C}$ such that $x \in C \subseteq U$ (by hypothesis), meaning $U \in \Tau'$ . Now suppose $W \in \Tau'$ . Then $W$ is expressible as the arbitrary union of elements of $\mathcal{C}$ , which are all elements of $\Tau$ , and so $W \in \Tau$ . Thus $\Tau = \Tau'$ .

It is difficult to describe topologies by their elements (as a topology is a subset of a power set, which is unwieldy). Thus we look towards a topological space's basis, a typically smaller subset of the topology that generates it under unions. We investigate properties of topologies by looking at bases of a topology.

[2.2.8]Theorem#

Suppose $X$ is some set with topologies $\Tau, \Tau'$ , each with a basis $\mathcal{B}, \mathcal{B}'$ respectively. Then the following statements are equivalent.

$\Tau$ is coarser than $\Tau'$ .
For each $x \in X$ and for every $B \in \mathcal{B}$ such that $x \in B$ , there exists some $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$ .

Proof.

$(2) \implies (1)$ : We seek to show $\Tau \subseteq \Tau'$ . Suppose $U \in \Tau$ . Then, since $\mathcal{B}$ generates $\Tau$ , there is some $B \in \mathcal{B}$ such that $x \in \mathcal{B} \subseteq U$ . By hypothesis, however, there is some $B' \in \Tau'$ such that $x \in \mathcal{B}' \subseteq U$ . Thus $\mathcal{B}'$ generates $\Tau$ , meaning $U \subseteq \Tau'$ .

$(1) \implies (2)$ : Suppose $x \in X$ is such that there is some $B \in \mathcal{B}$ for which $x \in B$ . Note $\mathcal{B} \subseteq \Tau \subseteq \Tau'$ , and so $B \in \Tau'$ . Since $\Tau'$ is generated by $\mathcal{B}'$ , there is some $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$ .

[2.2.9]Example(Topologies on

\bR

Consider the following topoologies that can be endowed on $\bR$ .

Let $\mathcal{B}$ be the set of all open intervals in $\bR$ .
$\mathcal{B} = \{(a, b): a < b\} = \{x \in \bR: a < x < b\}.$
The topology generated by $\mathcal{B}$ is said to be the standard topology on $\bR$ . All open sets of $\bR$ with respect to this topology take the form of the union of arbitrarily many open intervals.
Now let $\mathcal{B}$ be defined as such.
$\mathcal{B} = \{[a, b): a < b\} = \{x \in \bR: a \le x < b\}.$
The topology generated by $\mathcal{B}$ is said to be the lower limit topology. We denote $\bR$ endowed with this topology as $\bR_\ell$ .
Finally, define $K = \{1/n\}_{n \in \bZ_+}$ , and let $\mathcal{B}$ be defined as such.
$\mathcal{B} = \{(a, b), (a, b) \setminus K\}.$
The topology generated by $\mathcal{B}$ is said to be the K-topology. We denote $\bR$ endowed with this topology as $\bR_K$ .

[2.2.10]Remark#

The topologies on $\bR_\ell$ and $\bR_K$ (denoted $\Tau_\ell$ and $\Tau_K$ ) are both strictly finer than the topology on $\bR$ (denoted $\Tau$ ) and both incomparable.

Proof.

First, we show $\Tau_\ell$ is strictly finer than $\Tau$ . Let $x \in (a, b)$ , with $(a, b)$ an arbitrary basis element of $\Tau$ . Then note $x \in [x, b) \subseteq (a, b)$ , where $[x, b)$ is a basis element of $\Tau_\ell$ , and so $\Tau \subseteq \Tau_\ell$ ( $\Tau_\ell$ is finer than $\Tau$ ). To show that this is a strict subset, we show $\Tau_\ell \not \subseteq \Tau$ . Consider some $x \in [x, d)$ , with $[x, d)$ a basis element of $\Tau_\ell$ . Then note there is no open interval that contains $x$ and is a subset of $[x, d)$ . Thus $\Tau$ is not finer than $\Tau_\ell$ , meaning $\Tau_\ell$ is strictly finer than $\Tau$ .

Next, we show $\Tau_K$ is strictly finer than $\Tau$ . Note the basis of $\Tau_K$ comprises the basis of $\Tau$ , so it is automatically finer than $\Tau$ —we seek to show $\Tau$ is not finer than $\Tau_K$ . Let $0 \in (-1, 1) \setminus K$ , where $(-1, 1) \setminus K$ is a basis element of $\Tau_K$ . Then note there is no local refinement in $\Tau$ that contains $0$ (that is, there is no open interval that contains $0$ and is a subset of this basis element). Thus $\Tau$ is not finer than $\Tau_K$ , meaning $\Tau_K$ is indeed strictly finer than $\Tau$ .

Finally, we show $\Tau_\ell$ and $\Tau_K$ are incomparable (i.e., neither is finer than one another). Consider $2 \in [2, 3)$ , where $[2, 3)$ is a basis element of $\Tau_\ell$ . Then note no open interval $(a, b)$ —much less any punctured open interval $(a, b) \setminus K$ —contains $2$ and is a subset of $[2, 3)$ . Thus $\Tau_K$ is not finer than $\Tau_\ell$ . Conversely, consider $0 \in (-1, 1) \setminus K$ , with $(-1, 1) \setminus K$ a basis element of $\Tau_K$ . No left closed, right open interval (i.e., basis elements of $\Tau_\ell$ ) contains $0$ and is a subset of $(-1, 1) \setminus K$ . Thus $\Tau_\ell$ is not finer than $\Tau_K$ , completing the proof.

Consider some arbitrary topological space $(X, \Tau)$ , and let $\mathcal{B}$ be a basis for $\Tau$ . Recall that $\mathcal{B}$ must be a cover for $X$ , but it also carries the extra requirement that the intersection of two basis elements must comprise another basis element. In other words, a basis is closed under finite intersection. Then, what is the minimal structure that, when considering the set of all finite intersections, regenerates the basis? We define the subbasis and explain the hierarchy accordingly.

[2.2.11]Definition(Subbasis)#

A subbasis $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$ . The topology generated by the subbasis $\mathcal{S}$ is said to be the collection $\Tau$ of all unions of fintie intersections of elements of $\mathcal{S}$ .

[2.2.12]Remark(Interlude on Subbasis)#

Before we show the topology generated by a subbasis is, indeed, a topology, we provide some form of intuition for it.

A subbasis $\mathcal{S}$ is the most general collection of subsets capable of generating a topology. The primary distinction between a basis and a subbasis lies in the intersection axiom: while a basis $\mathcal{B}$ must be “self-refining”—in the sense that the intersection of any two basis elements must be representable as a union of other basis elements—a subbasis $\mathcal{S}$ is subject to no such constraint.

To bridge this gap, the construction of the topology $\Tau$ proceeds in two distinct stages:

Generating the Basis: We first take the collection of all finite intersections of elements in $\mathcal{S}$ . This produces a new collection $\mathcal{B}$ which is, by construction, a basis. Because $\mathcal{B}$ is formed under finite intersections, it satisfies the “self-refining” property that $\mathcal{S}$ lacked.
Generating the Topology: We then take the collection of all arbitrary unions of elements in $\mathcal{B}$ . This final collection $\Tau$ now satisfies the remaining axioms of a topology.

Conceptually, this identifies the topology $\Tau$ as the collection of all arbitrary unions of finite intersections of elements in $\mathcal{S}$ . This is the coarsest/minimal topology containing $\mathcal{S}$ , allowing one to define a topology by simply declaring a specific collection of sets to be open, bypassing the need to verify that the initial collection satisfies any intersection properties itself.

[2.2.13]Theorem(Subbasis Generates a Topology)#

Let $X$ be an arbitrary set. The set of all arbitrary unions of the set of all finite intersections of a subbasis forms a topology on $X$ .

Proof.

Let $X$ be a set we seek to endow a topology on. We show that the set of all finite intersections of a subbasis forms a basis, and the theorem follows immediately (since we know a basis generates a topology by taking arbitrary unions). Suppose our subbasis $\mathcal{S}$ is written as

\mathcal{S} = \{S_\lambda\}_{\lambda \in \Lambda}.

We seek to show $\mathcal{B}$ , the set of all finite intersections of elements of $\mathcal{S}$ , forms a basis. Of course, $\mathcal{S} \subseteq \mathcal{B}$ , and $\mathcal{S}$ covers $X$ , so automatically $\mathcal{B}$ covers $X$ as well. Now let

B_1 = S_{\lambda_1} \cap \cdots \cap S_{\lambda_m} \qquad B_2 = S_{\lambda_1}' \cap \cdots \cap S_{\lambda_n}',

where $B_1, B_2 \in \mathcal{B}$ . To complete the proof, we show some finite intersection of $S$ -sets is a subset of $B_1 \cap B_2$ . Indeed,

B_3 = B_1 \cap B_2 = (S_{\lambda_1} \cap \cdots \cap S_{\lambda_m}) \cap (S_{\lambda_1}' \cap \cdots \cap S_{\lambda_n}')

is a finite intersection of sets (by basic properties of sets), so $B_3 \in \mathcal{B}$ . Thus, each $x \in (B_1 \cap B_2)$ is such that $x \in B_3 \subseteq (B_1 \cap B_2)$ , completing the proof.

2.2.1Exercises#

[2.2.14]Problem#

Let $(X, \Tau)$ be a topological space, with $A \subseteq X$ . Suppose, for each $x \in A$ , there is some open set $U_x$ such that $x \in U_x \subseteq A$ . Show $A$ is open.

Proof.

Define $O = \bigcup_{x \in A} U_x$ . Note that it is an arbitrary union of open sets, and so it is open. Immediately, since each $U_x \subseteq A$ , it follows that $O \subseteq A$ . But also, since $\{U_x\}_{x \in A}$ forms a cover for $A$ , it's true that $A \subseteq O$ . Thus $A = O$ , and so $A$ is open.

[2.2.15]Problem#

Suppose $\{\Tau_\alpha\}_{\alpha \in A}$ is a family of topologies on $X$ . Show $\bigcap_{\alpha \in A} \Tau_\alpha$ is a topology.

Solution.

Let $T = \bigcap_{\alpha \in A} \Tau_\alpha$ . Then $T$ is the set of all sets that were open in every topology in $\{T_\alpha\}_{\alpha \in A}$ . We note that $\emptyset$ and $X$ are open in each $T_\alpha$ ( $\alpha \in A$ ), and so they belong in $T$ . Now consider arbitrarily many elements $\{O_\lambda\}_{\lambda \in \Lambda}$ from $T$ . We note these elements are in every $\Tau_\alpha: \alpha \in A$ , so their union should also be in every $\Tau_\alpha: \alpha \in A$ . Thus the union is in $T$ as well. An analogous statement may be made about the intersection of finitely many elements, which completes the proof.

[2.2.16]Problem#

Suppose $\mathcal{A}$ is the basis for some topology on $X$ . Show that the topology $\mathcal{A}$ generates is the intersection of all topologies on $X$ that contain $\mathcal{A}$ .

Solution.

Suppose $\mathcal{A}$ generates a topology $T$ . Let $\{\Tau_\alpha\}_{\alpha \in A}$ be the family of topologies on $X$ such that $\mathcal{A} \subseteq \Tau_\alpha$ for each $\alpha \in A$ . Let $T' = \bigcap_{\alpha \in A} \Tau_\alpha$ —we show $T = T'$ . Of course, $T \in \{\Tau_\alpha\}_{\alpha \in A}$ , and so $T' \subseteq T$ . Now we seek to show $T \subseteq T'$ . Equivalently, we show that for any open set in $T$ , it follows that aforementioned open set is also in $\Tau_\alpha$ for every $\alpha \in A$ . Fix $\alpha \in A$ and $O \in T$ . Note $\mathcal{A}$ is a basis for $T$ , and so for each $x \in O$ , there is some $B_x \in \mathcal{A}$ such that $x \in B_x \subseteq O$ . Thus $O = \bigcup_{x \in O} B_x$ . But note $B_x \in \mathcal{A} \subseteq \Tau_\alpha$ for every $x \in O$ , and so the arbitrary union of each $B_x$ is in $\Tau_\alpha$ . Thus $O \in \Tau_\alpha$ , which completes the proof.

[2.2.17]Problem#

Let $\mathcal{B} = \{(a, b): a < b \in \bQ\}$ . Show $\mathcal{B}$ is a basis for the standard topology on $\bR$ .

Solution.

Let $O \subseteq \bR$ be open with respect to the standard topology. Fix $x \in O$ , and choose $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon) \subseteq O$ . By the density of $\bQ$ in $\bR$ , there is some $a < b \in \bQ$ such that

x - \epsilon < a < x < b < x + \epsilon.

Then note $(a, b) = B \in \mathcal{B}$ , and thus $x \in B \subseteq (x - \epsilon, x + \epsilon) \subseteq O$ . Thus $\mathcal{B}$ is a basis for the standard topology on $\bR$ .