
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:

1. Every element $B \in \mathcal{B}$ is open in $X$.

2. 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. Omitted. $\square$

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. Omitted. $\square$

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. Omitted. $\square$

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

1. for $U \in \mathcal{B}$ the image $f(U)$ is open,

2. for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$, and

3. 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. $\square$

Comment #2590 by Dario Weißmann on

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

Comment #2787 by Shuddhodan K V on

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

Comment #2788 by Dario Weißmann on

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

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 on

@#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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