Criterion in terms of the bases for determining whether one topology is finer than another
When topologies are given by bases, one has a useful criterion in terms of the bases for determining whether one topology is finer than another:
Lemma 13.3. (Munkres's Topology p.81) Let $\mathscr{B}$ and $\mathscr{B}^{\prime}$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}^{\prime}$, respectively, on $X$. Then the following are equivalent:
(1) $\mathcal{T}^{\prime}$ is finer than $\mathcal{T}$.
(2) For each $x \in X$ and each basis element $B \in \mathscr{B}$ containing $x$, there is a basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $x \in B^{\prime} \subset B$.
If one modifies condition (2) as:
(2') For each nonempty basis element $B \in \mathscr{B}$ there is a nonempty basis element $B^{\prime} \in \mathscr{B}^{\prime}$ such that $B^{\prime} \subset B$.
then it is unlikely that one can conclude (2'): one cannot use the original argument for showing the implication (2) $\Rightarrow$ (1) anymore.
Can anyone give a counter-example for (2') $\Rightarrow$ (1)?
[(2') edited.]
1 answer
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| User | Comment | Date |
|---|---|---|
| Snoopy | (no comment) | Feb 4, 2023 at 17:24 |
A counterexample would be the set $S=\{0,1,2\}$ with the topologies $\mathcal T = \{\emptyset, \{0\}, \{0,1\}, S\}$and $\mathcal T' = \{\emptyset,\{0\},S\}$. Clearly $\mathcal T'$ is not finer than $\mathcal T$.
Now $\mathcal T$ is generated by the basis $\mathscr B = \{\{0\},\{0,1\},S\}$ and $\mathcal T'$ is generated by $\mathscr B' = \{\{0\},S\}$.
Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ is such that $B'\subset B$.
Thus this gives a counterexample to (2')$\implies$(1).
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