Definition talk:Proper Subset

Proper subset
Do we really want to require a proper subset to be non-empty? --Dfeuer (talk) 04:59, 29 December 2012 (UTC)


 * Look at the "Also defined as" and we see that this has been covered. We're back to the old issue of having picked a particular published definition which differs from somebody else's favourite definition. If we choose the definition which includes the empty set, then someone else is going to say, "Everybody knows a proper subset can't be empty, you fool!" And there it stands. --prime mover (talk) 09:15, 29 December 2012 (UTC)


 * This one causes real problems. See, e.g., Set Difference with Proper Subset. I know you are not fond of anything changing, but I think it would be very helpful to change this one... --Dfeuer (talk) 03:37, 4 March 2013 (UTC)


 * Doesn't that result hold whether the proper subset is empty or not? --GFauxPas (talk) 04:21, 4 March 2013 (UTC)


 * Indeed. But the way the page uses "proper subset" is not the way it is defined here. I frankly doubt that many pages actually do use it the way it is defined here. --Dfeuer (talk) 04:38, 4 March 2013 (UTC)


 * You seem to be suggesting the awkward of the two changes (the other one being changing the wrongly referring page), based on nothing but a hunch and your own limited experience. I for one am inclined to use "proper" as it is currently defined. I am wholly against changing the definition - in fact, I'm not even going to argue my case for a change. Please leave it alone. &mdash; Lord_Farin (talk) 08:26, 4 March 2013 (UTC)


 * Interestingly, I fail to see what exactly was wrong with the page. We are perfectly allowed to delay the conclusion "contradiction" once we've reached it. &mdash; Lord_Farin (talk) 08:29, 4 March 2013 (UTC)


 * We can always rewrite the proof of Set Difference with Proper Subset to say "Let $T$ be a set such that $T \subsetneq S$" and then the precise definition of "proper subset" would be less of an issue. Although I would be unwilling to rename the page because it would make it overly messy and obscure its basic simplicity. --prime mover (talk) 08:38, 4 March 2013 (UTC)


 * I repeat, there is no problem. In case $T$ is empty, we could infer a contradiction from $\forall x \in S: x \in T$, but it is not mandatory. In fact, the decision was made to defer that a bit because a general contradiction could also be established. &mdash; Lord_Farin (talk) 08:42, 4 March 2013 (UTC)