Prove that there is a unique topology is the coarsest containing topologies TαTαT_\alpha

{Tα}{Tα}\{T_\alpha\} is a family of topologies on XXX. Prove that there is a unique
coarsest topology containing topologies TαT_\alpha and a unique finest topology contained in all topologies TαT_\alpha.

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Using common terminology, the statement should read “prove that there is a unique topology, the coarsest, containing all topologies TaT_a and a unique topology, the finest, contained in all topologies TaT_aâ€‌.
– egreg
Oct 20 at 17:20

  

 

@MathematicsStudent1122: I’m pretty sure that the OP used a general bilingual dictionary to translate the terms that came out as rude and thin; in some non-mathematical contexts these are synonyms of the desired coarse and fine.
– Brian M. Scott
Oct 20 at 17:20

  

 

@lily Are you familiar with the fact that for every subset V⊆℘(X)\mathcal V\subseteq\wp(X) there exist a smallest topology that contains V\mathcal V (the topology generated by V\mathcal V)? So also for V:=⋃α∈ATα\mathcal V:=\bigcup_{\alpha\in A}T_{\alpha}.
– drhab
Oct 20 at 17:23

  

 

Ok. Then the finest topology contained in all topologies is the union and the intersection is the coarsest topology? @drhab
– lily
Oct 20 at 17:28

  

 

@lily I am afraid not. An intersection of topologies on the same set XX is a topology itself. To find the coarsest topology that contains every TαT_{\alpha} you must take the intersection of all topologies that contain ⋃α∈ATα\bigcup_{\alpha\in A}T_{\alpha}. The finest topology contained in every TαT_{\alpha} is intersection ⋂α∈ATα\bigcap_{\alpha\in A}T_{\alpha}. As said: both are topologies since they are both an intersection of topologies.
– drhab
Oct 20 at 17:41

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1 Answer
1

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On the second part:

Let τ\tau denote a topology such that it is contained in every TαT_{\alpha}.

So denoting the index set by AA we have:τ⊆⋂α∈ATα\tau\subseteq\bigcap_{\alpha\in A} T_{\alpha}\tag1

Fortunately it can be shown that ⋂α∈ATα\bigcap_{\alpha\in A} T_{\alpha} is a topology itself (try to prove that yourself) and (1)(1) tells us that any topology that is contained in every TαT_{\alpha} is coarser (i.e. has no more elements).

This allows us to conclude that ⋂α∈ATα\bigcap_{\alpha\in A} T_{\alpha} is the finest topology that is contained in every TαT_{\alpha}.

On the first part.

As said in the comments the intersection of a collection of topologies on XX is a topology itself.

If V⊆℘(X)\mathcal V\subseteq\wp(X) then we can take the collection of all topologies that contain V⊆℘(X)\mathcal V\subseteq\wp(X) and then take the intersection of these topologies. Denoting this intersection by τV\tau_{\mathcal V} we observe that:

τV\tau_{\mathcal V} is a topology.
V⊆τV\mathcal V\subseteq\tau_{\mathcal V}
If ρ\rho is a topology with V⊆ρ\mathcal V\subseteq\rho then τV⊆ρ\tau_{\mathcal V}\subseteq\rho

This together tells us that τV\tau_{\mathcal V} is the coarsest topology that contains V\mathcal V.

You are searching for the coarsest topology that contains every TαT_{\alpha} or equivalently for the coarsest topology that contains ⋃α∈ATα\bigcup_{\alpha\in A}T_{\alpha}, so you can apply this.