(In reply to
attempt at the rest by Charlie)
continuing:
e & f I've already stated were true.
g. The only subset of the null set is itself. It's not a proper subset as it has everything the null set has. It does not have the set consisting of the null set, so g is False.
h. seems to be true. Nothing is more restrictive than the null set, so ANDing leaves just the null set.
I. I don't think this is true. The null set does not contain a set consisting of the null set.
j. False. Not every set (arbitrarily chosen A) contains the null set, though the null set is a subset of every set,it's not a member of every set.
k. Since the null set is a subset of every set, the set of subsets of any set includes the null set, so the null set is indeed a member of omega. This one is True.
l. The same doesn't apply to l. The set consisting of the null set is not a subset of any arbitrary set A. This is false.
m. True, Uniting the null set to any given set does not add anything to what it didn't already have as a subset.
n. Already stated as True.
o. The set on the right has only one element. It has only two subsets: the null set and that complicated inner set. The set consisting of the null set is not one of these two. the statement is false.
I'm probably wrong on one or more of these as it gets complicated with different terminology for the same thing: phi, null set or {}. But I tried to be consistent in treating the null set, {} differently from the set containing only the null set, {{}}.

Posted by Charlie
on 20140419 14:49:15 