1
$\begingroup$

Any collection of topologies has a greatest lower bound among all topologies: $$\inf_{i\in I}\mathcal{T}_i=\bigcap_{i\in I}\mathcal{T}_i\text{ for }I\neq\{\}\\ \inf_{i\in I}\mathcal{T}_i=\mathcal{P}(X)\text{ for }I=\{\}$$ Especially that means: $$\inf_{i\in I}\mathcal{T}_i\text{ always exists}$$ Since $\inf A=\sup A_-$ and $\sup A=\inf A_+$ (see Suprema vs Infima) also: $$\sup_{i\in I}\mathcal{T}_i\text{ always exists}$$ Thus any collection admits a coarsest topology: $$\mathcal{T}_\mathcal{A}:=\inf_{\mathcal{A}\subseteq\mathcal{T}}\mathcal{T}:\quad \mathcal{A}\subseteq\mathcal{T}_\mathcal{A}$$ That is precisely saying the infimum was a minimum: $$\inf_{\mathcal{A}\subseteq\mathcal{T}}\mathcal{T}=\min_{\mathcal{A}\subseteq\mathcal{T}}\mathcal{T}$$

Assuming only collections containing the chaotic topology: $\{\varnothing,X\}\subseteq\mathcal{A}$

Does the same hold true mutatis mutandum for coarsest topology being replaced by finest topology?

$\endgroup$
  • 2
    $\begingroup$ No. For example if the family does not contain the empty set. $\endgroup$ – Rudy the Reindeer Jun 19 '14 at 6:52
  • $\begingroup$ Ok and what if the family contains at least the empty set and the space itself: $\varnothing,X\in\mathcal{A}$ $\endgroup$ – C-Star-W-Star Jun 19 '14 at 10:06
  • $\begingroup$ Then it depends on whether the smallest generated topology is a subset of the given family or not. Maybe you could add the new assumption to the question and then someone more knowledgable than myself will write an answer. $\endgroup$ – Rudy the Reindeer Jun 19 '14 at 11:30
  • 2
    $\begingroup$ (Correction of earlier comment): Suppose $X$ is the set in which we are working, and suppose $\mathcal{A} = \{\emptyset,A,B,X\}$ where $A,B$ are disjoint. Then $\{\emptyset,A,X\}$ and $\{\emptyset,B,X\}$ are distinct finest topologies in $\mathcal{A}$. This shows that there cannot be a unique finest topology, if that was part of your question. $\endgroup$ – Lee Mosher Jun 19 '14 at 16:45
  • $\begingroup$ Cool so that's it $\endgroup$ – C-Star-W-Star Jun 20 '14 at 0:03
0
$\begingroup$

This answer is due to Lee Mosher. Thank you alot!

Suppose neither $A\nsubseteq B$ nor $B\nsubseteq A$ and consider the collections $\mathcal{A}=\{\varnothing,A,B,X\}$. Then: $$\sup_{\mathcal{A}\supseteq\mathcal{T}}\mathcal{T}=\{\varnothing,A,B,A\cap B,X\}\nsubseteq\mathcal{A}$$

This also illuminates what usually happens: $$\{\varnothing,X\}\subseteq\mathcal{A},\{\varnothing,A,X\}\subseteq\mathcal{A},\{\varnothing,B,X\}\subseteq\mathcal{A}$$ So the supremum was no maximum!

Similarly we have for the empty collection $\mathcal{A}=\{\}$: $$\sup_{\mathcal{A}\subseteq\mathcal{T}}=\{\varnothing,X\}\nsubseteq\mathcal{A}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.