What results from the reaction of 5-bromopentan-1-ol and NaOH? In our chemistry class, we noted three molecules that result from the reaction:


pentane-(1,5)-diol




This one is simple to understand, simply a nucleophilic substitution between $\ce{Br^-}$ and $\ce{OH^-}$.


(1,5)-epoxypentane




This is the main result, resulting from the intramolecular esterification of the two $\ce{OH^-}$ groups of the previous molecule. This is wrong, please see the explanation 
by  ringo


4-penten-1-ol




This one, I don't understand. How is this a possible result? Could anyone help me understand? Thank you for your time.
The three reactions are:

An $\mathrm{S_N2}$:



Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S_N2}$):



An $\mathrm{E2}$: