Math4201 Topology I (Lecture 3)

Let $X=\mathbb{R}$ and $\mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).

Check properties 1:

for any $x\in \mathbb{R}$, $\exists (x-1,x+1)\in \mathcal{B}$ such that $x\in (x-1,x+1)$

Check properties 2:

let $B_1=(a,b)$ and $B_2=(c,d)$ be two basis elements, and $x\in B_1\cap B_2=(\max(a,c),\min(b,d))\in \mathcal{B}$.

Let $X=\mathbb{R}$ and $\mathcal{B}_{LL}=\{[a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).

Check properties 1:

for any $x\in \mathbb{R}$, $\exists [x,x+1)\in \mathcal{B}_{LL}$ such that $x\in [x,x+1)$

Check properties 2:

let $B_1=[a,b)$ and $B_2=[c,d)$ be two basis elements, and $x\in B_1\cap B_2=[max(a,c),min(b,d))\in \mathcal{B}_{LL}$.

Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of rectangle of the form $(a,b)\times (c,d)$ where $a,b,c,d\in \mathbb{R}$ and $a<b,c<d$. (boundary is not included)

Check properties 1:

for any $(x,y)\in \mathbb{R}^2$, $\exists (x,y)\in \mathcal{B}$ such that $(x,y)\in (x,y)$

Check properties 2:

let $B_1=(a,b)\times (c,d)$ and $B_2=(e,f)\times (g,h)$ be two basis elements, and $(x,y)\in B_1\cap B_2=(max(a,e),min(b,f))\times (max(c,g),min(d,h))\in \mathcal{B}$.

Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of open disks.

Check properties 1:

for any $x\in \mathbb{R}^2$, $\exists B_1(x)\in \mathcal{B}$ such that $x\in B_1(x)$.

Check properties 2:

let $B_{r_1}(x)$ and $B_{r_2}(y)$ be two basis elements, for every $z\in B_{r_1}(x)\cap B_{r_2}(y)$, $\exists B_{r_3}(z)\in \mathcal{B}$ such that $z\in B_{r_3}(z)\subseteq B_{r_1}(x)\cap B_{r_2}(y)$.

(even $B_{r_1}(x)\cap B_{r_2}(y)\notin \mathcal{B}$)

$\mathcal{T}_{\mathcal{B}}$ is a topology on $X$ because:

1. $\emptyset \in \mathcal{T}_{\mathcal{B}}$ because $\emptyset \in \mathcal{B}$. $X\in \mathcal{T}_{\mathcal{B}}$ because $\forall x\in X, \exists B\in \mathcal{B}$ such that $x\in B\subseteq X$ (by definition of basis (property 1)))

2. $\mathcal{T}_{\mathcal{B}}$ is closed under arbitrary unions.

   Want to show $\{U_\alpha | U_\alpha\in \mathcal{T}_{\mathcal{B}}\}_{\alpha \in I}\implies \bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}_{\mathcal{B}}$.

   Because $\forall x\in \bigcup_{\alpha \in I} U_\alpha$, $\exists \alpha_0$ such that $x\in U_{\alpha_0}$. Since $U_{\alpha_0}\in \mathcal{T}_{\mathcal{B}}$, $\exists B\in \mathcal{B}$ such that $x\in B\subseteq U_{\alpha_0}\subseteq \bigcup_{\alpha \in I} U_\alpha$.

3. $\mathcal{T}_{\mathcal{B}}$ is closed under finite intersections.

   Want to show $U_1,U_2,\ldots,U_n\in \mathcal{T}_{\mathcal{B}}\implies \bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$.

   If $n=2$, since $\forall x\in U_1\cap U_2$, $x\in U_1$ and $x\in U_2$, $\exists B_1\in \mathcal{B}$ such that $x\in B_1\subseteq U_1$ and $\exists B_2\in \mathcal{B}$.

   Applying the second property of basis, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2\subseteq U_1\cap U_2$.

   By induction, we can show that $\bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$.

Let $X$ be arbitrary.

Let $\mathcal{B}=\{x|x\in X\}$ (collection of all singleton subsets of $X$).

Then $\mathcal{T}$ is the discrete topology.

$(\Rightarrow)$

If $U\in \mathcal{T}_{\mathcal{B}}$, we want to show that $U$ is a union of basis elements.

For any $x\in U$, by the definition of $\mathcal{T}_{\mathcal{B}}$, there is a basis element $B_x$ such that $x\in B_x\subseteq U$.

So, $U\subseteq \bigcup_{x\in U} B_x$.

Since $\forall B_x\in \{B_x\}_{\alpha \in I}$, $B_x\subseteq U$, we have $U\supseteq\bigcup_{x\in U} B_x$.

So, $U=\bigcup_{x\in U} B_x$.

$(\Leftarrow)$

Applies observation 2.