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Q&A Criterion in terms of the bases for determining whether one topology is finer than another

posted 2y ago by celtschk‭ · edited 2y ago by celtschk‭

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#3: Post edited by $user avatar$ celtschk‭ · 2023-02-04T07:17:03Z (about 2 years ago)
fixed some grammar errors

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A counterexample would be the set $S = {0, 1, 2}$ with the topologies $T = {\emptyset, {0}, {0, 1}, S}$ and $T^{'} = {\emptyset, {0}, S}$ . Clearly $T^{'}$ is not finer than $T$ .
~~Now $T$ is generated by the basis $B = {{0}, {0, 1}, S}$ and $T^{'}$ is generated $B^{'} = {{0}, S}$ .~~
~~Now (2') is fulfilled, as for every non-empty $B \in B$ , the set $B^{'} = {0} \in B^{'}$ we have $B^{'} \subset B$ .~~
Thus this gives a counterexample to (2') $⟹$ (1).

A counterexample would be the set $S = {0, 1, 2}$ with the topologies $T = {\emptyset, {0}, {0, 1}, S}$ and $T^{'} = {\emptyset, {0}, S}$ . Clearly $T^{'}$ is not finer than $T$ .
Now $T$ is generated by the basis $B = {{0}, {0, 1}, S}$ and $T^{'}$ is generated by $B^{'} = {{0}, S}$ .
Now (2') is fulfilled, as for every non-empty $B \in B$ , the set $B^{'} = {0} \in B^{'}$ is such that $B^{'} \subset B$ .
Thus this gives a counterexample to (2') $⟹$ (1).

#2: Post edited by $user avatar$ celtschk‭ · 2023-02-04T07:14:10Z (about 2 years ago)
fixed typo

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A counterexample would be the set $S = {0, 1, 2}$ with the topologies $T = {\emptyset, {0}, {0, 1}, S}$ and $T^{'} = {\emptyset, {0}, S}$ . Clearly $T^{'}$ is not finer than $T$ .
~~Nor $T$ is generated by the basis $B = {{0}, {0, 1}, S}$ and $T^{'}$ is generated $B^{'} = {{0}, S}$ .~~
Now (2') is fulfilled, as for every non-empty $B \in B$ , the set $B^{'} = {0} \in B^{'}$ we have $B^{'} \subset B$ .
Thus this gives a counterexample to (2') $⟹$ (1).

A counterexample would be the set $S = {0, 1, 2}$ with the topologies $T = {\emptyset, {0}, {0, 1}, S}$ and $T^{'} = {\emptyset, {0}, S}$ . Clearly $T^{'}$ is not finer than $T$ .
Now $T$ is generated by the basis $B = {{0}, {0, 1}, S}$ and $T^{'}$ is generated $B^{'} = {{0}, S}$ .
Now (2') is fulfilled, as for every non-empty $B \in B$ , the set $B^{'} = {0} \in B^{'}$ we have $B^{'} \subset B$ .
Thus this gives a counterexample to (2') $⟹$ (1).

#1: Initial revision by $user avatar$ celtschk‭ · 2023-02-04T07:13:08Z (about 2 years ago)

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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).

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