# Finer and coarser topologies

We say that one topology \( \mathcal{T} \) on \( X \) is
*finer* than another topology \( \mathcal{T}' \) on \( X \)
iff every member of \( \mathcal{T}' \) is a member of \( \mathcal{T} \).
\( \mathcal{T}' \) is said to be *coarser* than
\( \mathcal{T} \).

*Stronger* and *weaker* are alternative
terminology for finer and coarser.

\( \mathcal{T} \) is *strictly* finer iff \( \mathcal{T}' \) is a proper subset.

Symbolically,

- \( \mathcal{T} \) is finer than \( \mathcal{T}' \) ⟺ \( \mathcal{T}' \subseteq \mathcal{T} \).
- \( \mathcal{T} \) is strictly finer than \( \mathcal{T}' \) ⟺ \( \mathcal{T}' \subset \mathcal{T} \).

#### Lemma. A basis for every basis.

Let \( X \) be a set. Let \( \mathcal{T} \) and \( \mathcal{T}' \) be topologies on \( X \) with bases \( \mathscr{B} \) and \( \mathscr{B}' \) respectively. Then \( \mathcal{T} \) is finer than \( \mathcal{T}' \) iff

for each \( x \in X \) and each basis element \( B' \in \mathscr{B'} \) containing \( x \), there is a basis element \( B \in \mathscr{B} \) 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 \( b' \in \mathscr{B'} \) can be expressed as a union of elements \( b \in \mathscr{B} \).

### Example

For a set \( X \), the discrete topology is the finest topology while the indiscrete topology is the coarsest topology.