NoteNextra-origin/content/Math4201/Math4201_L5.md

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.