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This question already has an answer here:

My book "Introduction to SetTheory" says

$\{\emptyset\}\in\{\{\emptyset\}\}$ but $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$

When we say $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean that $\emptyset$ is not listed amongst the elements of $\{\{\emptyset\}\}$.

I was under the impression that every set contains the empty set. Hence, the elements of $\{\{\emptyset\}\}$ would be $\emptyset,\{\emptyset\}$. If that were true, wouldn't the assertion that $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ be false?

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marked as duplicate by user91500, Vlad, Shahab, Stefan Mesken, Claude Leibovici Mar 24 '16 at 6:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ While it is true that for every set $E$, $\emptyset \subseteq E$, it is not true, in general, that $\emptyset \in E$. $\endgroup$ – Guest Mar 8 '14 at 2:14
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    $\begingroup$ The empty set is a subset of every set. But this does not mean it is an element of every set. $\endgroup$ – Chris Leary Mar 8 '14 at 2:17
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    $\begingroup$ In this context, "contains" means "has as a subset", not "has as an element". $\endgroup$ – MPW Mar 8 '14 at 2:17
  • $\begingroup$ In set theory the word "contains" is ambiguous and perhaps best avoided if you are just beginning to study the subject. $\endgroup$ – David Mar 8 '14 at 2:29
  • $\begingroup$ This math.stackexchange.com/questions/527819/… will help you. $\endgroup$ – Michael Hoppe Mar 8 '14 at 13:42
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$\varnothing\ne\{\varnothing\}$. The second is the only element of $\{\{\varnothing\}\}$. Do you see $\varnothing$ by itself listed in there? No.

When we say every set contains the empty set, we mean as a subset, not as a member or element, so we are saying that $\varnothing\subseteq X$ not $\varnothing\in X$. Do you think $\varnothing\in\varnothing$ for instance? That's not possible!

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The word contains is tricky, as it can mean both has as an element and is a superset of, depending on the context. Let's avoid it here.

Every set $A$ is a superset of $\emptyset$. But this is a theorem, not an axiom: there is a reason why this is true.

The definition of the subset relation is:

$B$ is a subset of $A$ (written $B\subseteq A$) whenever the following holds: $$ \forall x \, \big(x\in B\Rightarrow x\in A \big)$$

Note that this is trivially satisfied when $B=\emptyset$, since $\forall x \, (x\notin \emptyset)$. So the empty set is a subset of any $A$ because all of the elements of the empty set are members of $A$.

Now, in this case, we have two sets: $A=\{\emptyset\}$ and $B=\{\{\emptyset\}\}$. Each of these sets has one element. In the case of $A$, that element is the empty set. In the case of $B$, that element is a set with a single element, that single element being $\emptyset$. Since the set whose only element is the empty set is the set $A$, it follows that $B=\{A\}$.

Now, when we put it like that, it is pretty clear that $A\in B$. But why is $A$ not a subset of $B$? Well, for $A$ to be a subset of $B$, it would need to be the case that every element of $A$ is an element of $B$. The set $A$ has a single element, the empty set. So $A$ is a subset of $B$ if and only if $\emptyset\in B$.

But as we have seen, $B$ has only one element, and that element is $A$, not $\emptyset$ (and we know that $A\neq \emptyset$ since $A$ is a set with one element and $\emptyset is a set with no elements).

So, since $\emptyset\in A$ but $\emptyset \notin B$, it follows that $A\not\subseteq B$.

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  • $\begingroup$ Good point, contains is ambiguous and harmful to this discussion. $\endgroup$ – Pete Mar 8 '14 at 2:40
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Let's write $A = \{\emptyset\}$ and $B = \{\{\emptyset\}\}$. We are interested in whether or not $A \subseteq B$. In order to show that $A \subseteq B$, we would have to show that every member of $A$ is also a member of $B$. But $A$ has one member (namely $\emptyset$) which is not a member of $B$. So, it's not true that $A \subseteq B$.

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Hence, the elements of $\{\{\varnothing\}\}$ would be $\varnothing,\{\varnothing\}$.

No. $\{\{\varnothing\}\}$ contains only one element — $\{\varnothing\}$.
And like all sets with one element it has two subsets: $\{\{\varnothing\}\}$ and $\varnothing$.

Suppose, $A=\{B\}$. That means, $A$ contains one element — $B$ and two subsets — $A$ and $\varnothing$.
Suppose, $B=\{C\}$. Then $B$ contains one element (it is $C$), but that doesn't affect $A$ — it contains only one element either.

When you say $X\subseteq Y$ you mean that if $x\in X$ that $x\in Y$. In my example $C\in B$ but $C\not\in A$, so $B\not\subseteq A$. But $B\in A$.

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Neither of these is the empty set. On the left you have a set that contains the empty set as an element. The empty set is both an element and a subset of this set. On the right you have a set that, contains the set that contains the empty set as an element, as an element. The more you talk about it the more confusing it becomes. This is one of those things that once you meditate on it a bit it becomes obvious.

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