Source of most content: Gunhee Cho and Lecture note by Matsumura
Basis and Topology
Def. Given a set $X \ne \emptyset$, we call $\mathcal{B} (\subseteq 2^X)$ the basis on $X$ if it satisifes:
- $\mathcal{B}$ covers $X$ $\Leftrightarrow \forall x \in X, \exists B \in \mathcal{B}$ s.t. $x \in B$ $\Leftrightarrow X = \cup_{B \in \mathcal{B}} B$.
- If $x \in B_1 \cap B_2$ for $x \in X, \ B_1, B_2 \in \mathcal{B}$, then $\exists B_3 \in \mathcal{B}$ s.t. $x \in B_3 \subseteq B_1 \cap B_2$.
We call an element of basis, basis element and it is open.
Def. Given a set $X$, let $\mathcal{B}$ be a basis of $X$. We call $U$ is open in $X$ if $\forall x \in U, \exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subseteq U$, i.e., $U = \cup_{x \in U} B_x$. Thus, a topology generated by $\mathcal{B}$ is defined as $T_{\mathcal{B}} = \{ U \subseteq X : \forall x \in U, \exists B \in \mathcal{B} \text{ s.t. } x \in B \subseteq U\}$.
‘Topology’ generated by basis
Here, prove that topology on set $X$ generated by basis $\mathcal{B}$ is, in fact, topology.
Proof
- $\emptyset \in T_{\mathcal{B}}$ holds vacuously and $X \in T_{\mathcal{B}}$ since $\mathcal{B}$ covers $X$.
- Consider a collection of subsets $ U_{\alpha} \in T_{\mathcal{B}}, \forall \alpha \in I $. Then, we need to show $U := \cup_{\alpha \in I} U_{\alpha} \in T_{\mathcal{B}}$. Take any $x \in U$, then there exists $U_\alpha$ s.t. $x \in U_\alpha$. Since $U_\alpha \in T_{\mathcal{B}}$, there exists $B\in \mathcal{B}$ s.t. $x \in B \subseteq U_\alpha$. Since $U_\alpha \subseteq U$, this implies there exists $B \in \mathcal{B}$ s.t. $x \in B \subseteq U$, thus $U \in T_{\mathcal{B}} $.
- Consider a finite number of subsets $U_1, \cdots, U_n \in T_{\mathcal{B}}$. Then, we need to show $U := \cap_{i=1}^n U_i \in T_{\mathcal{B}}$. Then, prove by M.I.
- When $n=1$, it is trivial.
- Suppose it holds for $n-1$ subsets, i.e., $U_1 \cap \cdots \cap U_{n-1} \in T_{\mathcal{B}}$. Then, we can write $U’ = U_1 \cap \cdots \cap U_{n-1} \in T_{\mathcal{B}} $, i.e., $U = U’ \cap U_n$. Take any $x \in U$, then $x \in U’$ and $u\in U_n$, i.e., $\exists B_1, B_2 \in \mathcal{B}$ s.t. $x \in B_1 \subseteq U’$ and $x \in B_2 \subseteq U_n$. Since $\mathcal{B}$ is a basis, $\exists B_3 \in \mathcal{B}$ s.t. $x \in B_3 \subseteq B_1 \cap B_2$. Thus, $ x \in B_3 \subseteq B_1 \cap B_2 \subseteq U’ \cap U_n = U$, i.e., $U \in T_{\mathcal{B}}$.
\[\text{Let } (X,T) \text{ be a topological space (T.S). Let } \mathcal{B} \subseteq T. \text{ Then } \mathcal{B} \text{ is a basis and } T_{\mathcal{B}} = T \text{ iff } T=\{ \cup_{B \in \mathcal{B'}}B: \mathcal{B'} \subseteq \mathcal{B} \}. \nonumber\]
Proof
- $(\Rightarrow)$ Let $T = \{\cup_{B \in \mathcal{B’}}: \mathcal{B’} \subseteq \mathcal{B}\}$.
- If $U \in T_{\mathcal{B}}$, then $\forall x \in U, \exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subseteq U$, i.e., $U = \cup_{u\in U}B_u$. Therefore, $U \in T$, i.e., $T_{\mathcal{B}} \subseteq T$.
- If $U \in T$, then $U$ is a union of basis elements. Since basis elements are in $T_{\mathcal{B}}$, i.e., open, union of open sets is also open, thus $T \subseteq T_{\mathcal{B}}$.
Thus, $T = T_{\mathcal{B}}$.
- $(\Leftarrow)$
- (Basis-1) Since $X \in T$, we can write $X = \cup_{\alpha} B_{\alpha}$. Thus, $\forall x \in X, \exists B_\alpha$ s.t. $x \in B_\alpha$, i.e., $\mathcal{B}$ covers $X$.
- (Basis-2) If $B_1, B_2 \in \mathcal{B}$, then $B_1 \cap B_2 \in T$ since $B_1, B_2$ are open in $(X,T)$. Then, $B_1 \cap B_2 = \cup_{\alpha}B_{\alpha}$ for $B_{\alpha} \in \mathcal{B}$. Thus, $\forall x \in B_1 \cap B_2, \ \exists B_\alpha \in \mathcal{B}$ s.t. $x \in B_\alpha$. Here, $B_\alpha \subseteq B_1 \cap B_2$, i.e., $\mathcal{B}$ is a basis.
- (Equality-1) By definition of $T_{\mathcal{B}}$, if $U \in T_{\mathcal{B}}$ then $\forall x \in U, \exists B_x \in \mathcal{B}$ s.t. $x \in B_x \subseteq U$. Therefore, $U = \cup_{x \in U}B_u$. Since $B_x \in T$, $U$ is also open, i.e., $T_{\mathcal{B}} \subseteq T$.
- (Equality-2) If $U \in T$, then $U = \cup_{\alpha}B_{\alpha}$ for $B_{\alpha} \in \mathcal{B}$, i.e., $\forall x \in U,\exists B_\alpha$ s.t. $x \in B_\alpha$. $\implies U \in T_{\mathcal{T}}$, i.e., $T \subseteq T_{\mathcal{B}}$.
\[\text{Let } (X,T) \text{ be T.S and } \mathcal{B} \subseteq T. \text{ Then } \mathcal{B} \text{ is a basis and } T_{\mathcal{B}} = T \text{ iff } T=\{ U \subseteq X : \forall x \in U, \exists B_x \in \mathcal{B} \text{ s.t. } x \in B_x \subseteq U\}. \nonumber\]
Proof
- $(\Rightarrow)$ Trivial by definition of $T_{\mathcal{B}}$.
- $(\Leftarrow)$
- (Basis-1) By definition of $T$, $\mathcal{B}$ covers $X$.
- (Basis-2) Let $B_1, B_2 \in \mathcal{B} \subseteq T$. Then $B_1 \cap B_2 \in T$. By definition of $T$, $\mathcal{B}$ is a basis.
- (Equality-1) Same as the above proof (Equality-1) which implies $T_{\mathcal{B}} \subseteq T$.
- (Equality-2) Assumption implies $T \subseteq T_{\mathcal{B}}$.
NB1; From above results we can see that $T_{\mathcal{B}} = \{ U \subseteq X : \forall x \in U, \exists B_x \in \mathcal{B} \text{ s.t. } x \in B_x \subseteq U\} = \{ \cup_{B \in \mathcal{B’}}B: \mathcal{B’} \subseteq \mathcal{B} \}$.
\[\text{Given a set } X, \text{ Let } \mathcal{B}, \mathcal{B}' \text{ be a basis on X. Then, } T_{\mathcal{B}} \subseteq T_{\mathcal{B}'} \text{ iff } \forall B \in \mathcal{B}, \forall x \in B, \exists B_x' \in \mathcal{B}' \text{ s.t. } x \in B_x' \subseteq B. \nonumber\]
Proof
- $(\Rightarrow)$ Let $T_{\mathcal{B}} \subseteq T_{\mathcal{B}’}$ and take any $B \in \mathcal{A}$ and any $x \in B$. Since $x\in B$ where $B \in T_{\mathcal{B}} \subseteq T_{\mathcal{B}’}$, we can write $B=\cap_{x \in B}B_x’$. This implies $\exists x_0 \in B_{x_0}’$ s.t. $x_0 \in B_{x_0}’ \subseteq B$.
- $(\Leftarrow)$ Take any $U \in T_{\mathcal{B}}$. Then, we can write $U = \cup_{x \in U}B_x$. Since $\forall x \in U, B_x = \cup_{y \in B_x}B_y’$ by assumption, $U=\cup_{x\in U} ( \cup_{y \in B_x}B_{y}’ )$, i.e., $U$ is a union of basis elements in $\mathcal{B}’$. Therefore, $U \in T_{\mathcal{B}’}$, i.e., $T_{\mathcal{B}} \subseteq T_{\mathcal{B}’}$.
NB2; From this results, we can see that if $\mathcal{B} = \mathcal{B}’$ then $T_{\mathcal{B}} = T_{\mathcal{B}’}$.
Subbasis
Def. Given a set $X$, we call $S \subseteq 2^X$ s.t. $S = \{ S_i \} _{i \in I}$ where $S_i \subseteq X$ a subbasis of $X$ if $ \cup _{i \in I} S_i = X $.
Basis from Subbasis
\[\text{Let }S \text{ be a subbasis of }X. \text{ Then, } \mathcal{B} := \{ \cap_{\text{finite } i} S_i : S_i \in S, i\in I\} \text{ is a basis of } X. \nonumber\]Proof
- Since $X = \cup_{i \in I} S_i$, $\forall x \in X, \exists i \in I$ s.t. $x \in S_i$ where $S_i \in \mathcal{B}$.
- Let $x \in B_1 \cap B_2$ for $B_1, B_2 \in \mathcal{B}$. Then, we can write $B_1 = S_{i_1} \cap \cdots S_{i_n}, n < \infty$ and $B_2 = S_{j_1} \cap \cdots S_{j_m}, m < \infty$. Take $B_3 = S_{i_1} \cap \cdots S_{i_n} \cap S_{j_1} \cap \cdots S_{j_m}$, which is the intersection of two finite intersections, i.e., $B_3 \in \mathcal{B}$. Therefore, $x \in B_3 \subseteq B_1 \cap B_2$.
NB3; This implies that we can always generate the topology for any set $X$, since subbasis always exists.
NB4; Each basis element is open in $X$ w.r.t. $T_{\mathcal{B}}$. Also, the element of subbasis is open in $X$ as well.