Section 5.5 (004O): Bases—The Stacks project

5.5 Bases

Basic material on bases for topological spaces.

Definition 5.5.1. Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$ is called a base for the topology on $X$ or a basis for the topology on $X$ if the following conditions hold:

Every element $B \in \mathcal{B}$ is open in $X$ .
For every open $U \subset X$ and every $x \in U$ , there exists an element $B \in \mathcal{B}$ such that $x \in B \subset U$ .

The following lemma is sometimes used to define a topology.

Lemma 5.5.2. Let $X$ be a set and let $\mathcal{B}$ be a collection of subsets. Assume that $X = \bigcup _{B \in \mathcal{B}} B$ and that given $x \in B_1 \cap B_2$ with $B_1, B_2 \in \mathcal{B}$ there is a $B_3 \in \mathcal{B}$ with $x \in B_3 \subset B_1 \cap B_2$ . Then there is a unique topology on $X$ such that $\mathcal{B}$ is a basis for this topology.

Proof. Let $\sigma (\mathcal B)$ be the set of subsets of $X$ which can be written as unions of elements of $\mathcal B$ . We claim $\sigma (\mathcal{B})$ is a topology. Namely, the empty set is an element of $\sigma (\mathcal{B})$ (as an empty union) and $X$ is an element of $\sigma (\mathcal{B})$ (as the union of all elements of $\mathcal{B}$ ). It is clear that $\sigma (\mathcal{B})$ is preserved under unions. Finally, if $U, V \in \sigma (\mathcal{B})$ then write $U = \bigcup _{i \in I} U_ i$ and $V = \bigcup _{j \in J} V_ j$ with $U_ i, V_ j \in \mathcal{B}$ for all $i \in I$ and $j \in J$ . Then

$U \cap V = \bigcup \nolimits _{i \in I, j \in J} U_ i \cap V_ j$

The assumption in the lemma tells us that $U_ i \cap V_ j \in \sigma (\mathcal{B})$ hence we see that $U \cap V$ is too. Thus $\sigma (\mathcal{B}$ is a topology. Properties (1) and (2) of Definition 5.5.1 are immediate for this topology. To prove the uniqueness of this topology let $\mathcal T$ be a topology on $X$ such that $\mathcal B$ is a base for it. Then of course every element of $\mathcal{B}$ is in $\mathcal{T}$ by (1) of Definition 5.5.1 and hence $\sigma (\mathcal{B} \subset \mathcal{T}$ . Conversely, part (2) of Definition 5.5.1 tells us that every element of $\mathcal{T}$ is a union of elements of $\mathcal{B}$ , i.e., $\mathcal{T} \subset \sigma (\mathcal{B})$ . This finishes the proof. $\square$

Lemma 5.5.3. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$ . Let $\mathcal{U} : U = \bigcup _ i U_ i$ be an open covering of $U \subset X$ . There exists an open covering $U = \bigcup V_ j$ which is a refinement of $\mathcal{U}$ such that each $V_ j$ is an element of the basis $\mathcal{B}$ .

Proof. If $x \in U = \bigcup _{i\in I} U_ i$ , there is an $i_ x \in I$ such that $x \in U_{i_ x}$ . Thus we have a $B_{i_ x} \in \mathcal{B}$ verifying $x \in B_{i_ x} \subset U_{i_ x}$ . Set $J = \{ i_ x | x \in U\}$ and for $j = i_ x \in J$ set $V_ j = B_{i_ x}$ . This gives the desired open covering of $U$ by $\{ V_ j\} _{j \in J}$ . $\square$

Definition 5.5.4. Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$ is called a subbase for the topology on $X$ or a subbasis for the topology on $X$ if the finite intersections of elements of $\mathcal{B}$ form a basis for the topology on $X$ .

In particular every element of $\mathcal{B}$ is open.

Lemma 5.5.5. Let $X$ be a set. Given any collection $\mathcal{B}$ of subsets of $X$ there is a unique topology on $X$ such that $\mathcal{B}$ is a subbase for this topology.

Proof. By convention $\bigcap _\emptyset B = X$ . Thus we can apply Lemma 5.5.2 to the set of finite intersections of elements from $\mathcal B$ . $\square$

Lemma 5.5.6. Let $X$ be a topological space. Let $\mathcal{B}$ be a collection of opens of $X$ . Assume $X = \bigcup _{U \in \mathcal{B}} U$ and for $U, V \in \mathcal{B}$ we have $U \cap V = \bigcup _{W \in \mathcal{B}, W \subset U \cap V} W$ . Then there is a continuous map $f : X \to Y$ of topological spaces such that

for $U \in \mathcal{B}$ the image $f(U)$ is open,
for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$ , and
the opens $f(U)$ , $U \in \mathcal{B}$ form a basis for the topology on $Y$ .

Proof. Define an equivalence relation $\sim$ on points of $X$ by the rule

$x \sim y \Leftrightarrow (\forall U \in \mathcal{B} : x \in U \Leftrightarrow y \in U)$

Let $Y$ be the set of equivalence classes and $f : X \to Y$ the natural map. Part (2) holds by construction. The assumptions on $\mathcal{B}$ exactly mirror the assumptions in Lemma 5.5.2 on the set of subsets $f(U)$ , $U \in \mathcal{B}$ . Hence there is a unique topology on $Y$ such that (3) holds. Then (1) is clear as well. $\square$

Comments (6)

Comment #2590 by Dario Weißmann on June 02, 2017 at 23:27

Typo in Lemma 5.5.6: In the second sentence the "," sould be a "."

Comment #2621 by Johan on July 07, 2017 at 12:13

Thanks! Fixed here.

Comment #2787 by Shuddhodan K V on August 24, 2017 at 20:38

In Lemma 5.5.5 one should require $\cup_{B \in \mathcal{B}}B=X$ .

Comment #2788 by Dario Weißmann on August 27, 2017 at 21:03

I would say, that the empty intersection is the space $X$ . So the set of finite intersections contains $X$ and satisfies the conditions of lemma 5.5.2.

Comment #2790 by Fan Zheng on August 29, 2017 at 03:37

In Lemma 5.5.6, it seems easier to take $Y=X$ (as a set), $f=\text{id}_X$ and apply Lemma 5.5.2 with $\mathcal{B}$ to give the topology on $Y$ .

Comment #2895 by Johan on October 07, 2017 at 15:14

@#2790: Hmm, yes you are right. The construction as currently given has an additional property (namely that Y is Kolmogorov)... I'm going to leave it as is since it is rather trivial anyway.

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5.5 Bases

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