I know this statement
$$\{\varnothing \} \neq \varnothing$$
is true because { $\varnothing $} $\subsetneq \varnothing$ since $ \varnothing \notin \varnothing $. Is this true?
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2Latex touch: Prefer varnothing to emptyset which looks a little too skinny. – Jean Marie Mar 13 '17 at 23:33
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@JeanMarie I never noticed that until I saw your comment. I'll never use emptyset again! – The Count Mar 13 '17 at 23:53
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The set consists of exactly one set, namely the null set. So it is not empty; i.e. it is not the null set.
PossumP
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The null set indicates there is 1 element in there, thus it can't be empty. – Lily Mar 13 '17 at 23:28
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And so, this set with one element cannot be a subset of the null set (which is a set with no elements). I think your statement mixes the idea of "sets" with the idea of a "non-set." (The English is somewhat imprecise.) – PossumP Mar 13 '17 at 23:46
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Here is a correct proof of the statement. Denote by $\text{card}\,A$ the number of elements in a finite set $A$. We have $$ \text{card}\, \{\emptyset\} = 1, \quad \text{card}\, \emptyset = 0 $$ and $$ 1 \not = 0. $$ Therefore, because equal sets contain the same number of element, $\{\emptyset\} \not = \emptyset$.
Olivier
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Since the other answers give the other things I would have said, here's the third way to think about it I would list:
In your title, the first object (on the left) is a set consisting of elements which are themselves sets. The elements, of which there is only one, are all the empty sets.
The second object is a set which, by definition, consists of nothing.
EDIT: I just thought of this option as well. Since the empty set is the set of nothing, we may write $\varnothing=\{\;\}$. Then the answer is obvious.
The Count
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