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 $\mathcal{T}'$ )

Example (discrete topology is finer than the trivial topology)

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.

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'

Proof

$(\Rightarrow)$

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 $T'$ .

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 $\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$ .

Example (lower limit topology is strictly finer than the standard topology)

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.