# Fine toplogy

1. Aug 19, 2011

### attardnat

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

2. Aug 20, 2011

### 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.

3. Aug 20, 2011

### 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.

4. Aug 20, 2011

### 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?

5. Aug 21, 2011

### 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?