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$\{\emptyset\}$ is a set containing the empty set. Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?

My hypothesis is yes by looking at the form of "the superset $\{\{\emptyset\}\}$" which contains "the subset $\{\emptyset\}$".

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    $\begingroup$ No, there is only one element {0} in {{0}} $\endgroup$ – Shuchang Sep 12 '13 at 8:21
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    $\begingroup$ Is every element of the first set also an element of the last set? $\endgroup$ – Stefan Hansen Sep 12 '13 at 8:22
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    $\begingroup$ What you are really noticing when you look at the form is that "the set $\{\{\emptyset\}\}$" contains the element $\{\emptyset\}$. $\{\emptyset\}$ is an element of $\{\{\emptyset\}\}$, not a subset of it. $\endgroup$ – Ari Brodsky Sep 12 '13 at 13:27
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It's easier to just replace everything by variables, then it's clearer.

Set $a=\varnothing, b=\{\varnothing\}=\{a\}$ and $c=\{\{\varnothing\}\}=\{\{a\}\}=\{b\}$.

Now the question is whether or not $\{a\}\subseteq\{b\}$. But since $a\neq b$, it's easy to see that the answer is negative.

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    $\begingroup$ I would add for all clarity that $a\neq b$ follows from the facts that $\emptyset\in b$ while $\emptyset\notin a$. But the important step is indeed to reduce the question to whether $a=b$ or not. $\endgroup$ – Marc van Leeuwen Sep 12 '13 at 8:47
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    $\begingroup$ Marc, thank you for the excellent comment! $\endgroup$ – Asaf Karagila♦ Sep 12 '13 at 8:54
  • $\begingroup$ @MarcvanLeeuwen That would be called assuming the conclusion. The point is to prove that $\emptyset \notin a$, $\emptyset \in b$, and if you assume that to show that $a\ne b$, that defeats the whole point of making the second statement at all. $\endgroup$ – AJMansfield Sep 12 '13 at 21:36
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    $\begingroup$ @AJMansfield: I don't see what you could have given you the conclusion that I assumed anything. I just indicated that one can justify $a\neq b$ by pointing at an element of $b\setminus a$. One could go further and explain why $\emptyset\in a$ and $\emptyset\notin b$ by applying the defintion of $a,b$ respectively. But nowhere is there an assumption. $\endgroup$ – Marc van Leeuwen Sep 13 '13 at 3:50
  • $\begingroup$ My problem with this answer is that a is a subset of b, which in turn is a subset of c. Isn't it true that if a set contains a subset, all elements of that subset are in the set? $\endgroup$ – Max Oct 14 '18 at 19:10
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No, $\emptyset \in \{\emptyset\}$ but $\emptyset \notin \{\{\emptyset\}\}$.

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    $\begingroup$ @Neil I don't think so. It state: "Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?" $\endgroup$ – William Sep 12 '13 at 8:29
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    $\begingroup$ Wow! 6 upvotes for this! $\endgroup$ – William Sep 12 '13 at 8:43
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    $\begingroup$ William, there were over a 140 upvotes to Mariano's decomposition of $\pi$ into decimal digits, and over 200 (250?) to the infamous $\sf W$. What's 8 votes? :-) $\endgroup$ – Asaf Karagila♦ Sep 12 '13 at 8:55
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    $\begingroup$ @AsafKaragila, agreed, voting behavior on this site is pretty crazy. To quality of an answer is, in many ways, orthogonal to the number of votes it attracts (although positively related to its overall rank). $\endgroup$ – goblin GONE Sep 12 '13 at 11:00
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    $\begingroup$ The infamous $W$: math.stackexchange.com/questions/74347/… $\endgroup$ – Jean-Sébastien Sep 12 '13 at 15:18
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For such abstract questions, it is important that you stick to the definitions of the involved notions.

By definition, $A$ is a subset of $B$ if every element contained in $A$ is also contained in $B$.

Now we look at $A = \{\emptyset\}$ and $B = \{\{\emptyset\}\}$. $A$ has exactly one element, namely $\emptyset$. This is not an element of $B$, since the only element of $B$ is $\{\emptyset\}$ and $\emptyset \neq \{\emptyset\}$ (see below). So $A$ is not a subset of $B$.

Why $\emptyset \neq \{\emptyset\}$? By definition, two sets are equal if they contain the same elements. However, $\emptyset$ is the empty set without any element, but $\{\emptyset\}$ is a $1$-element set with the element $\emptyset$.

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  • $\begingroup$ They're often interchangeable, but can I suggest "every" rather than "any" in the first definition (so that "…if every element contained in $A$…)? In this case it's (to me, anyway) a little to easy to read it as "some" (as we might in "$A$ and $B$ have a common factor if any factor of $A$ is also a factor of $B$). $\endgroup$ – Joshua Taylor Sep 12 '13 at 15:07
  • $\begingroup$ That said, +1 because this answer appeals to the definitions of subset and set equality, which I find makes this clearer than the currently highest upvoted answer. $\endgroup$ – Joshua Taylor Sep 12 '13 at 15:08
  • $\begingroup$ @JoshuaTaylor: Thanks! I agree with your suggestion on "any" and "every" and modified my answer accordingly. $\endgroup$ – azimut Sep 12 '13 at 15:10
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{{∅}} has only one element, namely {∅}, it implies there are only the two trivial subsets, {{∅}} which contains all elements and ∅ which contains none.

formally all subsets of {{∅}}

{{∅}}⊆{{∅}}
∅⊆{{∅}}

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Think of {∅} as a box containing an empty box, then {{∅}} is a box containing a box which contains an empty box, in which case {{∅}} is box containing {∅}, which means

{∅} is subset of {{∅}}

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  • $\begingroup$ :) I know ... just having fun $\endgroup$ – Amit Kriplani Sep 12 '13 at 13:04
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    $\begingroup$ I think your analogy leads to: $$\{∅\} \text{ is an element of } \{\{∅\}\}.$$ A subset is equivalent to opening the bigger box, removing some of the contents (possibily all or none) and then closing it again. So, like other $1$ element sets, $\{\{∅\}\}$ has only two possible subsets: itself or the empty set. $\endgroup$ – Henry Sep 12 '13 at 13:08

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