# Math4201 Topology I (Lecture 5 Bonus)

## Comparison of two types of topologies

Let $X=\mathbb{R}^2$ and the two types of topologies are:

The "circular topology":

$$
\mathcal{T}_c=\{B_r(p)\mid p\in \mathbb{R}^2,r>0\}
$$

The "rectangle topology":

$$
\mathcal{T}_r=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}
$$

> Are these two topologies the same?

### Comparison of two topologies

#### Definition of finer and coarser

Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$. We say $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. We say $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$. We say $\mathcal{T}$ and $\mathcal{T}'$ are equivalent if $\mathcal{T}=\mathcal{T}'$.

$\mathcal{T}$ is strictly finer than $\mathcal{T}'$ if $\mathcal{T}'\subsetneq \mathcal{T}$. (that is, $\mathcal{T}'$ is finer and not equivalent to $\mathcal{T}$)
$\mathcal{T}$ is strictly coarser than $\mathcal{T}'$ if $\mathcal{T}\subsetneq \mathcal{T}'$. (that is, $\mathcal{T}$ is coarser and not equivalent to $\mathcal{T}'$)

<details>
<summary>Example (discrete topology is finer than the trivial topology)</summary>

Let $X$ be an arbitrary set. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$

The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$

Clearly, $\mathcal{T}_1 \subseteq \mathcal{T}_0$.

So the discrete topology is finer than the trivial topology.

</details>

#### Lemma

> [!TIP]
>
> Motivating condition:
>
> We want $U$ be an open set in $\mathcal{T}'$, then $U$ has to be open with respect to $\mathcal{T}$. In other words, $\forall x\in U, \exists$ some $B\in \mathcal{B}$ such that $x\in B\subseteq U$.

Let $\mathcal{T}$ and $\mathcal{T}'$ be topologies on $X$ associated with bases $\mathcal{B}$ and $\mathcal{B}'$. Then

$$
\mathcal{T}\text{ is finer than } \mathcal{T}'\iff \text{ for any } x\in X, x\in B'\in \mathcal{B}', \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq B'
$$

<details>
<summary>Proof</summary>

$(\Rightarrow)$

Let $B'\in \mathcal{B}'$. If $x\in B'$, then $B'\in \mathcal{T}'$ and $T$ is finer than $T'$, so $B'\in \mathcal{T}$.

Take $T=\mathcal{T}_{\mathcal{B}}$. $\exists B\in \mathcal{B}$ such that $x\in B\subseteq B'$.

$(\Leftarrow)$

Let $U\in \mathcal{T}$. Then $U=\bigcup_{\alpha \in I} B_\alpha'$ for some $\{B_\alpha'\}_{\alpha \in I}\subseteq \mathcal{B}'$.

For any $B_\alpha'$ and any $x\in \mathcal{B}_\alpha'$, $\exists B_\alpha\in \mathcal{B}$ such that $x\in B_\alpha\subseteq B_\alpha'$.

Then $B_\alpha'$ is open set in $\mathcal{T}$.

So $U$ is open in $\mathcal{T}$.

$T$ is finer than $T'$.

</details>

Back to the example:

For every point in open circle, we can find a rectangle that contains it.

For every point in open rectangle, we can find a circle that contains it.

So these two topologies are equivalent.

#### Standard topology in $\mathbb{R}^2$

The standard topology in $\mathbb{R}^2$ is the topology generated by the basis $\mathcal{B}_{st}=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}$. This is equivalent to the topology generated by the basis $\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$.

<details>

<summary>Example (lower limit topology is strictly finer than the standard topology)</summary>

The lower limit topology is the topology generated by the basis $\mathcal{B}_{ll}=\{[a,b)\mid a,b\in \mathbb{R},a<b\}$.

This is finer than the standard topology.

Since $(a,b)\in \mathcal{B}_{st}$, we have $\forall x\in (a,b), \exists B=[x,b)\in \mathcal{B}_{ll}$ such that $x\in B\subsetneq (a,b)$.

So the lower limit topology is strictly finer than the standard topology.

$[0,1)$ is not open in the standard topology. but it is open in the lower limit topology.

</details>