Source of most content: Gunhee Cho and Lecture note by Matsumura
Topology on $\mathbb{R}$
Def. Given a set $X=\mathbb{R}$, we call a topology $T_{std}$ generated by basis $\mathcal{B} = \{ (a,b) : a < b, a,b \in \mathbb{R} \}$ the standard topology on $\mathbb{R}$.
Def. Given a set $X=\mathbb{R}$, we call a topology $T_{l}$ generated by basis $\mathcal{B} = \{ [a,b) : a < b, a,b \in \mathbb{R} \}$ the lower limit topology on $\mathbb{R}$.
Def. Given a set $X=\mathbb{R}$ and let $K=\{\frac{1}{n} : n \in \mathbb{N}\}$, we call a topology $T_{K}$ generated by basis $\mathcal{B} = \{ (a,b), (a,b)-K : a < b, a,b \in \mathbb{R} \}$ the K-topology on $\mathbb{R}$.
Properties
- $T_{std} \subsetneq T_{l}$, i.e., lower limit topology is finer.
- $T_{std} \subsetneq T_{K}$, i.e., K-topology is finer.
- $T_{l}$ and $T_{K}$ are not comparable.
Proof
- 1 and 2 can be easily proved from relation between bases.
- $(0,1)\setminus \{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \} \not\in T_{l}$ and $[0,1) \not\in T_{K}$.
Some induced topologies
Def. Given two T.S. $(X, T_X)$ and $(Y, T_Y)$, then $\mathcal{B} := \{ U_X \times V_Y : U_X \in T_X, V_Y \in T_Y \} \subseteq 2^{X\times Y}$ forms a basis of topology and the topology generated by $\mathcal{B}$ is called box topology on $X\times Y$.
Check
Here, show that above $\mathcal{B}$ is basis on $X\times Y$.
- Take any $(x,y) \in X\times Y$. Since $X$ and $Y$ are open in $X$ and $Y$, respectively, thus $(x,y) \in X\times Y \in \mathcal{B}$.
- Let $(x,y)\in B_1 \cap B_2$ for $B_1, B_2 \in \mathcal{B}$. Write $B_1 = U_X^1 \times V_Y^1$ and $B_2 = U_X^2 \times V_Y^2$. Let $B_3 :=B_1 \cap B_2 = ( U_X^1 \times V_Y^1) \cap (U_X^2 \times V_Y^2) = (U_X^1 \cap U_X^2) \times (V_Y^1 \cap V_Y^2)$. Thus, $(x,y) \in B_3 \subseteq B_1 \cap B_2$.
Proposition
\[\text{Let } \mathcal{B}_X, \mathcal{B}_Y \text{ be a basis of } (X, T_X), (Y, T_Y) \text{, respectively. Then, } \mathcal{B}_X \times \mathcal{B}_Y \text{ be a basis of the box toplogy on } X\times Y \nonumber\]Proof
Let $T_{\mathcal{B}}$ be a toplogy generated by a basis $\mathcal{B} := \mathcal{B}_X \times \mathcal{B}_Y = \{ B_x \times B_y : B_x \in \mathcal{B}_X, B_y \in \mathcal{B}_Y\}$ on $X \times Y$, i.e., $T _{\mathcal{B}} = \{ W \subseteq X \times Y : \forall (x,y) \in W, \exists B \in \mathcal{B} \text{ s.t. }(x,y) \in B \subseteq U \}$. Take $(x,y) \in W$. Then, there are $B = B_X \times B_Y$ for $B_x \in \mathcal{B}_X$ and $B_y \in \mathcal{B}_Y $ s.t. $(x,y) \in B$. Since $B_x$ and $B_y$ are bases of $T_X$ and $T_Y$, respectively, there exist $U \in T_X$ and $V \in T_Y$ s.t. $x \in B_x \subseteq U$ and $y \in B_y \subseteq V$. Thus, it shows that $\mathcal{B}_X \times \mathcal{B}_Y$ is a basis of the topology on $X\times Y$ and $T _{\mathcal{B}}$ is the box toplogy on $X\times Y$ by NB1.
Def. Given T.S. $(X, T_X)$ and let $A \subseteq X$. Then, $T_A := \{ U \cap A : U \in T_X \}$ is a topology on $A$ and called the subspace topology.
Check
Here, show that $T_A$ is a topology.
- $\emptyset = \emptyset \cap A \in T_A$ and $A = A \cap A \in T_A$.
- Given $(U_i \cap A)_{i \in J}$, $\bigcup _{i \in J} (U_i \cap A) = (\bigcup _{i \in J} U_i) \cap A$. Since $U_i$ is open in $T_X$ for all $i\in J$, $(\bigcup _{i \in J} U_i)$ is also open. Thus, $(\bigcup _{i \in J} U_i) \cap A \in T_A$.
- Given $(U_i \cup A)_{i \in J}$ for $| J | < \infty$, $\bigcap _{i \in J} (U_i \cup A) = (\bigcap _{i \in J} U_i) \cup A$. Similarly to 2, $(\bigcap _{i \in J} U_i) \cup A \in T_A$.
Lemma
\[\text{Let } \mathcal{B} \text{ be a basis of } (X, T). \text{ For } A \subseteq X, \text{ then } \mathcal{B}_A := \{ B \cap A : B \in \mathcal{B} \} \text{ is a basis of subspace topology } T_A=T_{\mathcal{B}_A}. \nonumber\]Proof
By definition, $\mathcal{B}_A \subseteq T_A$. Given open set $U \cap A \in T_A$, take $x \in U \cap A$. Then, $\exists B \in \mathcal{B}$ s.t. $x \in B \subseteq U$ since $\mathcal{B}$ generates $T$. Then, by $B \cap A \in \mathcal{B}_A$, we have $x \in B\cap A \subseteq U \cap A$.
Proposition
\[\text{Given T.S.} (X, T_X), (Y, T_Y) \text{ and } A \subseteq X, B \subseteq Y. \text{ Then, } T_{box} \text{ on } A\times B = \text{ subspace topology } T_{A\times B}. \nonumber\]NB1; This means that (1) box and then subspace is equivalent to (2) subspace and then box.
Proof
Let $\mathcal{B}_X$ and $\mathcal{B}_Y$ be bases of $(X, T_X), (Y, T_Y)$, respectively.
- $\mathcal{B}_A = \{ C \cap A : C \in \mathcal{B}_X \}$ and $\mathcal{B}_B = \{ C \cap B : D \in \mathcal{B}_Y \}$. Then, $\mathcal{B} _{box} = \mathcal{B}_A \times \mathcal{B}_B = \{B_A \times B_B : B_A = C \cap A, B_B = D \cap B, C \in \mathcal{B}_X, D \in \mathcal{B}_Y\}$.
- $\mathcal{B}_{X \times Y} = \{C \times D : C \in \mathcal{B}_X, D \in \mathcal{B}_Y\}$. Then, basis of subspace, $\mathcal{B} _{A\times B} = \{C\cap A \times D \cap B : C \in \mathcal{B}_X, D \in \mathcal{B}_Y\} = \{B_A \times B_B : B_A = C \cap A, B_B = D \cap B, C \in \mathcal{B}_X, D \in \mathcal{B}_Y\}$.
Since bases of (1) and (2) are the same, topologies generated by these bases are same by NB2.