Generating Topology from a Basis
Buildup
Topology
For a set $X$, a collection $\mathscr{T}$ of subsets of $X$ that satisfies the following three conditions is called the topology on set $X$.
- $(T1)$ $\varnothing, X \in \mathscr{T}$
- If $(T2)$ $U_{\alpha} \in \mathscr{T} (\alpha \in \Lambda)$, then $\bigcup_{\alpha \in \Lambda} U_{\alpha} \in \mathscr{T}$.
- If $(T3)$ $U_{1},\cdots,U_{n} \in \mathscr{T}$, then $\bigcap_{i=1}^{n}U_{i} \in \mathscr{T}$.
In simple terms, a collection of subsets that contains the empty set and the whole set and is closed under union and countable intersection is called a topology.
Basis
For a set $X$, a collection $\mathscr{B}$ of subsets of $X$ that satisfies the following two conditions is called the basis of the topology on $X$. To avoid confusion with the basis in linear algebra, it is usually simply referred to as the basis of set $X$.
- $(B1)$ For every point $x\in X$, there exists a $B\in \mathscr{B}$ that satisfies $x\in B$. In other words, $\bigcup_{B\in \mathscr{B}}B=X$ holds.
- $(B2)$ For any $B_{1}, B_2 \in \mathscr{B}$ and a point $x \in (B_{1}\cap B_{2})$, there exists a $B_{3} \in \mathscr{B}$ that satisfies $x\in B_{3} \subset (B_{1}\cap B_{2})$.
Definition
Let $\mathscr{B}$ be called the basis of set $X$. A collection $\mathscr{T}_\mathscr{B}$ of subsets $U$ of $X$ that satisfies the following condition is called the topology on $X$ generated by $\mathscr{B}$. $$ \forall x \in U,\ \exists B\in \mathscr{B}\quad \text{s.t.}\ x\in B \subset U $$ In other words, $$ \mathscr{T}_\mathscr{B} =\left\{ U\subset X \ :\ \forall x \in U,\ \exists B\in \mathscr{B}\quad \text{s.t.}\ x\in B \subset U\right\} $$
Theorem
- $(0)$: $\mathscr{T}_\mathscr{B}$ is a topology on $X$.
Let $\mathscr{B}$ be said to be the basis of $X$. Then, $\mathscr{T}_\mathscr{B}$ has the following properties.
- $(a1)$: $\mathscr{T}_{\mathscr{B}}$ is equivalent to the collection of unions of elements of $\mathscr{B}$. $$ \mathscr{T}_{\mathscr{B}}=\left\{ \bigcup_{B\in \mathscr{{B}}^{\ast}}B\ :\ \mathscr{B}^{\ast}\subset \mathscr{B} \right\} $$
- $(b1)$ $\mathscr{T}_{\mathscr{B}}$ is the smallest topology on $X$ that includes $\mathscr{B}$.
Given a topological space $(X, \mathscr{T})$, the following two facts are equivalent about $\mathscr{B} \subset \mathscr{T}$.
- $(a2)$: $\mathscr{B}$ is the basis of $\mathscr{T}$. In other words, $\mathscr{T}=\mathscr{T}_{\mathscr{B}}$.
- $(b2)$: For any open set $U \in \mathscr{T}$ and point $x \in U$, there exists a $B\in \mathscr{B}$ that satisfies $x \in B \subset U$.
Explanation
That $\mathscr{B}$ is the basis of the topological space $(X,\mathscr{T})$ implies $\mathscr{T}=\mathscr{T}_\mathscr{B}$. It can be verified that $\mathscr{T}_\mathscr{B}$ actually becomes the topology on $X$.
Depending on the textbook, $(a1)$ can be introduced as the definition of the basis $\mathscr{B}$.
$(a2)$ and $(b2)$ deal with finding the basis from the given topology. On top of that, it covers how to generate a topology from the given basis, while below, it discusses how to find the basis that generates the topology when a topology is given.
This grounds for adopting the collection of open balls as the basis on Euclidean space $\mathbb{R}^n$.
Proof
$(0)$
$(T1)$
Because there does not exist a $x$ that satisfies $x\in \varnothing$, it follows that $\varnothing \in \mathscr{T}_\mathscr{B}$. Since $\mathscr{B}$ is the basis of $X$, by $(B1)$, for any point $x\in X$, there exists a $B \in \mathscr{B}$ that satisfies $x \in B \subset X$. Therefore, $X \in \mathscr{T}_\mathscr{B}$
$(T2)$
Let $U_\alpha \in \mathscr{T}_\mathscr{B}\ (\alpha \in \Lambda)$. For any point $x \in \bigcup_{\alpha \in \Lambda} U_\alpha$, there exists a $\alpha_{0} \in \Lambda$ such that $x \in U_{\alpha_{0}}$. Since $U_{\alpha_{0}} \in \mathscr{T}_{\mathscr{B}}$, by the-definition of $\mathscr{T}_{\mathscr{B}}$, there exists a $B \in \mathscr{B}$ that satisfies $x \in B \subset U_{\alpha_{0}}$. Therefore, $$ x \in B \subset U_{\alpha_{0}} \subset \bigcup_{\alpha \in \Lambda}U_{\alpha} $$ so $\bigcup_{\alpha \in \Lambda} U_{\alpha} \in \mathscr{T}_{\mathscr{B}}$. Let $(T3)$ $U,V \in \mathscr{T}_{\mathscr{B}}$. For any point $x \in U \cap V$, since $x\in U$ and $x \in V$, by the-definition of $\mathscr{T}_{\mathscr{B}}$, $$ x \in B_{1}\subset U,\quad x\in B_{2}\subset V $$ there exists a $B_{1}, B_{2}\in \mathscr{B}$ that satisfies it. Also, since $x\in B_{1}\cap B_2$, by $(B2)$ $$ x \in B_{3} \subset B_{1}\cap B_2 \subset U\cap V $$ there exists a $B_{3} \in \mathscr{B}$ that satisfies it. Therefore, by the definition of $\mathscr{T}_\mathscr{B}$, $U \cap V \in \mathscr{T}_\mathscr{B}$ is true.
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$(a1)$
Let $ \mathscr{T}_{\mathscr{B}} \supset \left\{ \bigcup_{B\in \mathscr{{B}}^{\ast}}B\ :\ \mathscr{B}^{\ast}\subset \mathscr{B} \right\} $$ \mathscr{B}$ be any subset $\mathscr{B}^{\ast} \subset \mathscr{B}$. If we denote by $U=\bigcup_{B\in \mathscr{B}^{\ast}} B$, then for any $x \in U$, there exists a $x\in B_{x} \in \mathscr{B}^{\ast}$. Thus, $x \in B_{x} \subset U$ leads to $U \in \mathscr{T}_{\mathscr{B}}$. If we denote by $ \mathscr{T}_{\mathscr{B}} \subset \left\{ \bigcup_{B\in \mathscr{{B}}^{\ast}}B\ :\ \mathscr{B}^{\ast}\subset \mathscr{B} \right\} $$ U \in \mathscr{T}_{\mathscr{B}}$, then by the-definition of $\mathscr{T}_{\mathscr{B}}$, for any $x \in U$, there exists a $B_{x} \in \mathscr{B}$ that satisfies $x \in B_{x} \subset U$. Therefore, since $U$ can be expressed as the union of elements of $\mathscr{B}$ as $U=\bigcup_{x\in U}B_{x}$, it follows that $U \in \left\{ \bigcup_{B\in \mathscr{{B}}^{\ast}}B\ :\ \mathscr{B}^{\ast}\subset \mathscr{B} \right\}$
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$(b1)$
First, show that $\mathscr{T}_{\mathscr{B}}$ includes $\mathscr{B}$. Let’s say $B \in \mathscr{B}$. Then, for $x\in B$, there exists a $B \in \mathscr{B}$ that satisfies $x \in B \subset B$, therefore $B \in \mathscr{T}_{\mathscr{B}}$. Now, consider the topology $\mathscr{T}$ on set $X$ that includes $\mathscr{B}$. If for $\mathscr{B}^{\ast} \subset \mathscr{B}$, $\mathscr{B}^{\ast} \subset \mathscr{B} \subset \mathscr{T}$ holds, by the-definition of topology $(T1)$, $\bigcup_{B \in \mathscr{B}^{\ast}} B$ also belongs to $\mathscr{T}$. Therefore, $\mathscr{T}_{\mathscr{B}} \subset \mathscr{T}$ follows. In other words, $\mathscr{T}_{\mathscr{B}}$ is the smallest topology that includes $\mathscr{B}$.
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$(a2) \iff (b2)$
$(a2) \implies (b2)$
Follows by the-definition of $\mathscr{T}_{B}$.
$(b2) \implies (a2)$
First, show that $\mathscr{B}$ becomes the basis of $X$.
- $(B1)$ Since the whole set $X$ is also an open set, for any point $x \in X$, there exists a $B \in \mathscr{B}$ that satisfies $x \in B \subset X$.
- $(B2)$ For any $B_{1}, B_2 \in \mathscr{B}$ and $x \in B_{1} \cap B_{2}$, since $\mathscr{B} \subset \mathscr{T}$, $B_{1}, B_{2} \in \mathscr{T}$ follows. By the-definition of topology $(T2)$, $B_{1} \cap B_2 \in \mathscr{T}$ holds. Also, by $(b2)$, there exists a $B_{3} \in \mathscr{B}$ that satisfies $x\in B_{3} \subset B_{1} \cap B_2$, fulfilling the condition of the basis.
Now, show that the topology generated by $\mathscr{B}$ is the same as $\mathscr{T}$. If we denote by $U\in \mathscr{T}$, then by $(b2)$, for each point $x \in U$, there exists a $B \in \mathscr{B}$ that satisfies $x \in B \subset U$. Therefore, by the definition of $\mathscr{T}_\mathscr{B}$, $\mathscr{T} \subset \mathscr{T}_\mathscr{B}$ holds. Since $(b1)$ implies that $\mathscr{T}_\mathscr{B}$ is the smallest topology that includes $\mathscr{B}$ and since $\mathscr{B} \subset \mathscr{T}$, $\mathscr{T}_\mathscr{B} \subset \mathscr{T}$ is true. Thus, $\mathscr{T}=\mathscr{T}_\mathscr{B}$ follows.
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