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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 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. $\square$

    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}

    Comments (5)

    Comment #2590 by Dario Weißmann on June 2, 2017 a 11:27 pm UTC

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

    Comment #2621 by Johan (site) on July 7, 2017 a 12:13 pm UTC

    Thanks! Fixed here.

    Comment #2787 by Shuddhodan K V on August 24, 2017 a 8:38 pm UTC

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

    Comment #2788 by Dario Weißmann on August 27, 2017 a 9:03 pm UTC

    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 a 3:37 am UTC

    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 $Y$.

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