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?

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    $\begingroup$ 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
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    $\begingroup$ (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

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 $\mathcal{A}=\{\}$: $$\sup_{\mathcal{A}\subseteq\mathcal{T}}=\{\varnothing,X\}\nsubseteq\mathcal{A}$$


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