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Supoose that we want to enumerate the elements contained in the below sets

(a) $\emptyset$ , (b) $\{\emptyset\}$ , (c) $\{\{\emptyset\}\}$, (d) $\{\emptyset, \{\emptyset\}\}$

The question is simple. How many elements contain each set?

My solution: The first set is the empty set so it contains 0 elements. The second set contains an empty set, consequently, contains 1 element. The third one contains 1 element because contains a set which set contains the empty set. Finally, the last one contains 2 elements. The empty set and the set that contains an empty set. Is this solution correct ?

My mind bleeds !!!

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    $\begingroup$ Yes. Your answers are correct. What helps for me is to have a very firm grasp that a collection (a set) as a specific thing, is a completely different thing and and different concept then the things within the collections as things. A registry of gym members is not the people who are members of the gym. And the reference books inside a reference library is not the library itself. $\endgroup$ – fleablood Nov 18 '18 at 19:23
  • $\begingroup$ I usually tell the students who struggle with all of the bracketing that unless it is the empty set, there is at least one element, so the number of elements is the number of commas +1. This is obviously not applicable to most sets, but it is a nice intuition builder. $\endgroup$ – Don Thousand Nov 18 '18 at 23:58
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For the sake of having an answer, yes your solution as well as your justifications are correct.

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    $\begingroup$ I think a comment would have been sufficient for this, not an answer. $\endgroup$ – Don Thousand Nov 18 '18 at 19:02
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    $\begingroup$ @RushabhMehta I guess it depends; having an upvoted answer means that the question can be removed from the unanswered list. $\endgroup$ – user279515 Nov 18 '18 at 19:54
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    $\begingroup$ then answer with more substance, ervx, not just "to answer". @fleabloods comment says far more than your "pseudo-answer". $\endgroup$ – amWhy Nov 18 '18 at 22:50
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    $\begingroup$ Relevant: How to answer proof-verification questions? $\endgroup$ – user279515 Nov 19 '18 at 5:19
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    $\begingroup$ @RushabhMehta: This answer is not even complete. The asker did not justify why (d) has exactly two elements. It has at most two, since there is only one comma, but why exactly two? $\endgroup$ – user21820 Nov 19 '18 at 17:14

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