Fine Topology on [0,1]: Equivalence to Euclidean Topology?

[attardnat]

Can anyone please help me with this because I'm really getting confused. Thanks!

In R, we know that fine topology is equivalent to the Euclidean topology as convex functions are continuous.

Now if instead of R we consider a subset of it say [0,1], the fine topology induced on [0,1] would it be equivalent to the Euclidean topology induced on [0,1] ?

Thanks once again

 

[Landau]

What do you mean by 'equivalent' topologies?

I am not familiar with the fine topology, but if by equivalent topologies you simply mean 'the same topology' (i.e. the same open set), then it is of course a tautology.

 

[attardnat]

yes i mean the same topology.
Convex functions on [0,1] are discontinuous at the boundaries so I don't understand how they generate the same topology as continuous functions.

 

[Landau]

Ah, so by 'induced on [0,1]' you don't mean the subspace topology. Could you define the fine topology for me? Is it the initial topology on X w.r.t. all convex functions X->R?

 

[attardnat]

I am not sure if i understood you well (as I'm not very much familiar with topology)

What I am trying to ask is the following: convex functions on R generate the fine topology and convex functions on R are the continuous functions so obviosly they generate the same topology. But since on [0,1], convex functions are not continuous, can they generate the same topology?