## Tag `004O`

## 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 abasis 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.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 asubbasis 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

- 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. $\square$

The code snippet corresponding to this tag is a part of the file `topology.tex` and is located in lines 224–332 (see updates for more information).

```
\section{Bases}
\label{section-bases}
\noindent
Basic material on bases for topological spaces.
\begin{definition}
\label{definition-base}
Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$
is called a {\it base for the topology on $X$} or a {\it basis for the
topology on $X$} if the following conditions hold:
\begin{enumerate}
\item Every element $B \in \mathcal{B}$ is open in $X$.
\item For every open $U \subset X$ and every $x \in U$,
there exists an element $B \in \mathcal{B}$ such that
$x \in B \subset U$.
\end{enumerate}
\end{definition}
\noindent
The following lemma is sometimes used to define a topology.
\begin{lemma}
\label{lemma-make-base}
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.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-refine-covering-basis}
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 $\mathcal{B}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-subbase}
Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$
is called a {\it subbase for the topology on $X$} or a {\it subbasis for the
topology on $X$} if the finite intersections of
elements of $\mathcal{B}$ form a basis for the topology on $X$.
\end{definition}
\noindent
In particular every element of $\mathcal{B}$ is open.
\begin{lemma}
\label{lemma-subbase}
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.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-create-map-from-subcollection}
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
\begin{enumerate}
\item for $U \in \mathcal{B}$ the image $f(U)$ is open,
\item for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$, and
\item the opens $f(U)$, $U \in \mathcal{B}$
form a basis for the topology on $Y$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-make-base}
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.
\end{proof}
```

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