Post History
#3: Post edited
by
celtschk
·
2023-02-04T07:17:03Z (almost 2 years ago)
fixed some grammar errors
- 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 $\mathscr B' = \{\{0\},S\}$.Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ we have $B'\subset B$.- Thus this gives a counterexample to (2')$\implies$(1).
- 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).
#2: Post edited
by
celtschk
·
2023-02-04T07:14:10Z (almost 2 years ago)
fixed typo
- 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 $\mathcal T$.
Nor $\mathcal T$ is generated by the basis $\mathscr B = \{\{0\},\{0,1\},S\}$ and $\mathcal T'$ is generated $\mathscr B' = \{\{0\},S\}$.- Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ we have $B'\subset B$.
- Thus this gives a counterexample to (2')$\implies$(1).
- 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 $\mathscr B' = \{\{0\},S\}$.
- Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ we have $B'\subset B$.
- Thus this gives a counterexample to (2')$\implies$(1).
#1: Initial revision
by
celtschk
·
2023-02-04T07:13:08Z (almost 2 years ago)
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 $\mathcal T$.
Nor $\mathcal T$ is generated by the basis $\mathscr B = \{\{0\},\{0,1\},S\}$ and $\mathcal T'$ is generated $\mathscr B' = \{\{0\},S\}$.
Now (2') is fulfilled, as for every non-empty $B\in\mathscr B$, the set $B'=\{0\}\in \mathscr B'$ we have $B'\subset B$.
Thus this gives a counterexample to (2')$\implies$(1).