Talk:Included Set Topology on Finite Intersection

This follows trivially from $A \subseteq B \implies \tau_B \subseteq \tau_A$ which is at least intuitively true. Probably we want that as a separate page. I suggest to name it something like 'Included Set Topology Reverses Subset' or 'Included Set Topology of Subset is Finer'. --Lord_Farin 12:02, 25 February 2012 (EST)


 * Yes we already have that result, it's Expansion of Included Set Topology, but I can't see how it follows trivially. --prime mover 16:44, 25 February 2012 (EST)

I retract the statement that this is trivial; some work is required (I have posted the proof). It seems I tacitly had the subset the other way around in my mind when I wrote the above. --Lord_Farin 17:49, 25 February 2012 (EST)


 * Yes, so did I when I initially worked it (see initial version of this proof page which was the first rubbish). I had vaguely pieced it together in my head the way you did it while I was having a nap this morning, but I lost it again after I'd had a day of domestic responsibilities and distractions and couldn't pick up the threads. So, thanks for that. Good luck with assignment. --prime mover 18:17, 25 February 2012 (EST)