I'm having a tough time understanding how the set theory of null sets work. I have: $$ X=\emptyset,\quad\quad Y = \{\emptyset\},\quad\quad Z = \{\{\emptyset\}\}. $$

Some of my self-study exercises include these true or false questions. Now, I'm more concerned with the reasoning behind why they're true or false as opposed to the answers as I already have the answers, I just want the understanding.

  1. $\emptyset \in X$.

    I know this is false because the null set is not an element of any set.

  2. $\emptyset \in Y$.

    I don't know why this is true.

  3. $\emptyset \in Z$.

    I don't know why this is false.

  4. $X \subseteq Y$.

    I know this is true because the null set symbol is directly within the set.

  5. $Y \subseteq Z$.

    I don't know why this is true.

  6. $X \in Y$.

    The same reason why (2) is true, I understand this one.

  7. $Y \in Z$.

    This is true because $\{\emptyset\}$ is directly within the set defined by $Z$.

  • 2
    $\begingroup$ Your reasoning for 1) is incorrect. Think of $\emptyset$ like an empty box. What is inside of an empty box? Well, there is nothing in there, so any statement of the form $blah \in \emptyset$ is false. $\endgroup$
    – Braindead
    May 14, 2015 at 5:36
  • 2
    $\begingroup$ For set $Y$. Imagine putting an empty box inside of another empty box. $Y$ is your "outer" box. What is inside of your outer box? Well, there is that empty box we put in there. Do you think you can continue from here? $\endgroup$
    – Braindead
    May 14, 2015 at 5:38
  • 1
    $\begingroup$ Your reasoning for 4) is incorrect. $\endgroup$
    – Braindead
    May 14, 2015 at 5:40
  • 1
    $\begingroup$ Out of curiosity, why did you think the null set can't be an element of any set? $\endgroup$
    – David Z
    May 14, 2015 at 17:58

5 Answers 5


Very important facts: Set membership ($\in$) and set inclusion ($\subseteq$) are two very different things. When we write $A = \{b\}$, it implies that $b \in A$ but it does not imply that $b \subseteq A$. The only way for $b \subseteq A$ to be true is if $b$ is a set and the set $b$ does not contain any members that are not also members of $A$.

Now to apply these facts:

  1. The null set can be an element of a set. (For example, it is an element of $Y$.) But the null set has no elements, and since $X=\emptyset$, $X$ has no elements and you cannot write $v \in X$ for any $v$ whatsoever, even $\emptyset$.

  2. $\emptyset \in Y$ because it was written that it is, as clearly as can be. The notation $Y=\{v\}$ means that $Y$ has one element, and $v$ is that element. Well, let $v=\emptyset$, that is, $Y=\{\emptyset\}$. The statement we made before about $v$ is now true about $\emptyset$: $Y$ has one element, and $\emptyset$ is that element.

  3. $Z$ has just one element. That element is the set $\{\emptyset\}$. But $\emptyset\neq\{\emptyset\}$. Therefore $\emptyset$ is not an element of $Z$.

  4. It is true that $X \subseteq Y$, but this is not because $\emptyset$ is an element of $Y$. It is because $X=\emptyset$, and $\emptyset$ is a subset of any other set that can ever be. In other words, it doesn't matter what is in $Y$.

  5. This is false. In fact, $Y \not\subseteq Z$, because $\emptyset \in Y$, but $\emptyset \not\in Z$. Again, $Z$ has just one element and that element is the set $\{\emptyset\}$, which is not the same thing as $\emptyset.$

  6. You are right, this is the same as 2. Since $X=\emptyset$, when we write $\emptyset\in Y$ we are saying that $X\in Y$.

  7. I don't know the mathematical definition of "directly within" as applied to sets. I suppose you mean that $\{\emptyset\}$ is found in the list of elements between the outer braces in the definition of $Z$, $Z=\{\ldots\}.$ So yes, $\{\emptyset\} \in Z$, and since $Y=\{\emptyset\}$, that implies $Y\in Z$.

  • $\begingroup$ @user153330: Says who? ZFCU is ZFC with urelements. NFU also has urelements. And nearly every mathematician does not think of $1,2,3$ as sets, nor do they think that it makes sense to ask whether $2 ⊆ \sqrt{-1}$ or not... In other words, there is nothing wrong with David K's very precise and robust statement that holds true no matter what foundational system we choose. In fact, foundational and philosophical issues aside, it is very good pedagogy to explicitly emphasize type-checking, because lack of type-checking is the cause of majority of mathematical confusion. $\endgroup$
    – user21820
    Feb 9, 2020 at 5:54

Here is a good "crutch" that you can use for these types of problems.

$X= \emptyset$ is an empty box, $Y = \{\emptyset\}$ is a box that contains an empty box, $Z = \{\{\emptyset\}\}$ is a box that contains a box that contains an empty box.

  1. Does an empty box contain an empty box? (No, an empty box is empty.)

  2. Does a box that contains an empty box contain an empty box? (Yes.)

  3. It is easy to get confused here. $Z$ contains a box that contains a box. $Z$ does not contain an empty box. In other words, we don't care that $Z$ contains a box that contains an empty box.

  4. Is everything inside the empty box also in $Y$? (Yes, because there is nothing inside the empty box.)

  5. Are all the elements in the box $Y$ in the box $Z$? (No, $Z$ does not contain an empty box).

  6. This is the same statement as 2.

  7. Does $Z$ contain a box that contains an empty box? Yes.

I find that sometimes thinking about sets in this manner provides a helpful paradigm shift.

  • $\begingroup$ I really doubt that the box analogy is helpful. For example if we have the sets $A=\{1,\{42\}\}$ and $B=\{2,\{42\}\}$, we would have to imagine a box containing the number $42$, and that box is inside both box $A$ and box $B$, but neither $A$ nor $B$ are inside each other. Boxes don't behave that way! $\endgroup$ May 14, 2015 at 19:11
  • 1
    $\begingroup$ What if we have disjoint instances of $\{42\}$? $\endgroup$
    – user157227
    May 17, 2015 at 19:40
  • 1
    $\begingroup$ x @user: We don't -- or at least we shouldn't, when we're trying to explain a set theory that satisfies the Axiom of Extensionality: There does not exist two different sets whose members are exactly the same. $\endgroup$ May 17, 2015 at 19:48
  • $\begingroup$ @HenningMakholm I have a solution that satisfies both Henning Makholm's and user157227's disagreement. Let $C$ be the box that contains $42$. Then at some time, say 1:00am, let $A$ be a box containing $1$ and $C$. Then at 2:00am, let $B$ be a box containing $2$ and $C$. Because $A$ and $B$ are defined at different times, it is possible to change ownership of $C$ while still maintaining the Axiom of Extensionality – that $C$ is a unique box. It's not a perfect analogy, because set theory is time-independent, but it might be useful for beginners. $\endgroup$
    – chharvey
    Feb 10, 2017 at 1:41

1)$X=\emptyset$ means that X is the null set that means it doesn't contain any elements or in a better way the subset of X is the null set.

2) $Y=\{\emptyset\}$ means that $\emptyset$ is an element of this set Y than means it contains this element $\emptyset$. You can also this this as that the subsets of this set will be the element and the null set

3) $Z=\{\{\emptyset\}\}$ means that this set $\{\emptyset\}$ is an element of the set $Z$ . This means that the subset will contain the set$\{\emptyset\}$ which is an element of $Z$ and the null set.

I think after this logic you will be able to answer all your questions


Others have already provided some good answers. This answer looks at determining set inclusion mechanistically, which may be easier to work with to get started.

Instead of writing the empty set as $\emptyset$, write it as {}. Then for each statement of the form $a \in B$, just look within the outermost set of braces of $B$ to see if you can find an instance of $a$. If it's there, then the statement is true. If it isn't, then the statement is false.

So, for example, the empty set {} doesn't contain the empty set. The set that does contain the empty set looks like {{}}.


I recently read a proof of why the null set is contained in every set, a concept which heretofore had confused me.

In formal logic, if the antecedent of an implication is false, then the implication as a whole is true. This is because the truth value of the consequent is independent of the truth value of the antecedent (though the converse of this statement is false).

With this in mind, consider the following implication:

$x$ $\in$ $\emptyset$ $\implies$ $x$ $\in$ $S$,

where $S$ is an arbitrary set. By the definition of the empty set, the antecedent is always false. Thus the implication itself is always true.

This is why your reasoning behind (1) is false. The reason "false" is the correct answer is because $X$ $=$ $\emptyset$; in other words, $X$ contains nothing. The null set is something, and so $X$ cannot contain it.

(2) is true because $Y$ clearly contains $\emptyset$.

(3) is technically true, for the argument given above. But suppose $Z$ is some exceptional, nonexistent set that does not contain the null set, which seems to be the assumption of whoever wrote this problem. The message being conveyed is that a set containing the null set is not the same as the null set itself, which is why it is false, even though such a set does not, and cannot, actually exist.

(4) is true, but vacuously so. This is because $X$ $\subseteq$ $Y$ can be rewritten as $x$ $\in$ $X$ $\implies$ $x$ $\in$ $Y$. As shown above, this implication is true because the antecedent is always false.

(5) is actually false. The notation here is a bit subtle. If $A$ and $B$ are arbitrary sets, then, if $A$ $\subseteq$ $B$, $x$ $\in$ $A$ $\iff$ $x$ $\in$ $B$, for all possible $x$. However, if $A$ $\subset$ $B$, $A$ is known as a "proper subset" of $B$, and implies that there exists at least one element $x$ such that $x$ $\in$ $B$ $\land$ $x$ $\notin$ $A$; in other words, all elements of $A$ are contained in $B$, though the converse of this statement is false. It may be the case that whoever wrote these problems uses $\subseteq$ to mean "proper subset of", in which case (5) is true.

(6) is true for the same reason (4) is false.

(7) is true for the same reason (6) is true.

It is important to make the distinction between $X$ $\subset$ $Y$, which means all elements of $X$ are contained in $Y$, and $X$ $\in$ $Y$, which means the set $X$ itself, though not necessarily the individual elements of $X$, is contained in $Y$.


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