Post History
#3: Post edited
- A counterexample would be the set
with the topologies and . Clearly is not finer than . Now is generated by the basis and is generated .Now (2') is fulfilled, as for every non-empty , the set we have .- Thus this gives a counterexample to (2')
(1).
- A counterexample would be the set
with the topologies and . Clearly is not finer than . - Now
is generated by the basis and is generated by . - Now (2') is fulfilled, as for every non-empty
, the set is such that . - Thus this gives a counterexample to (2')
(1).
#2: Post edited
- A counterexample would be the set
with the topologies and . Clearly is not finer than . Nor is generated by the basis and is generated .- Now (2') is fulfilled, as for every non-empty
, the set we have . - Thus this gives a counterexample to (2')
(1).
- A counterexample would be the set
with the topologies and . Clearly is not finer than . - Now
is generated by the basis and is generated . - Now (2') is fulfilled, as for every non-empty
, the set we have . - Thus this gives a counterexample to (2')
(1).
#1: Initial revision
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).