Math and science::Topology

Finer and coarser topologies

We say that one topology $T$ on $X$ is finer than another topology $T^{'}$ on $X$ iff every member of $T^{'}$ is a member of $T$ . $T^{'}$ is said to be coarser than $T$ .

Stronger and weaker are alternative terminology for finer and coarser.

$T$ is strictly finer iff $T^{'}$ is a proper subset.

Symbolically,

$T$ is finer than $T^{'}$ ⟺ $T^{'} \subseteq T$ .
$T$ is strictly finer than $T^{'}$ ⟺ $T^{'} \subset T$ .

Lemma. A basis for every basis.

Let $X$ be a set. Let $T$ and $T^{'}$ be topologies on $X$ with bases $B$ and $B^{'}$ respectively. Then $T$ is finer than $T^{'}$ iff

for each $x \in X$ and each basis element $B^{'} \in B^{'}$ containing $x$ , there is a basis element $B \in B$ such that $x \in B \subseteq B^{'}$ .

It should be easy to prove this. Refer to Munkres for a proof example.

By this lemma, it can be seen how a topology induced by one metric can be equivalent to a topology induced by another.

A basis for every basis: alternative perspective

I think a better way of phasing the second part of the iff relationship is to say: every $b^{'} \in B^{'}$ can be expressed as a union of elements $b \in B$ .

Example

For a set $X$ , the discrete topology is the finest topology while the indiscrete topology is the coarsest topology.

Finer and coarser topologies

Lemma. A basis for every basis.

A basis for every basis: alternative perspective

Example

Context