I have only recently been exposed to sets. According to Wikipedia, as seen on the first bullet mark of the link, $ \forall A: \emptyset \subseteq A$

Does this mean that $\emptyset$ is an element of all sets? (This is False, thank you those that answered)

Is the empty set also a set itself?

Assuming these statements are true, then the empty set therefore an element of the empty set. This does not sound right, please clarify for me. Thank you.

  • 7
    $\begingroup$ Being a subset and being an element are different. For every set $A$, $\varnothing\subseteq A$ but not necessarily $\varnothing\in A$. $\endgroup$ – Clayton Jan 8 '13 at 4:05
  • $\begingroup$ So an element is something without the {} curly brackets. And since $\emptyset$ has such brackets, it does not fit. $\endgroup$ – Leonardo Jan 8 '13 at 4:07
  • $\begingroup$ Well, something inside of { } means we're thinking of it as a set, that does not mean that something without the brackets is an element. By the looks of it, the notation seems to be the barrier here. $\endgroup$ – Andrew D Jan 8 '13 at 4:10
  • $\begingroup$ Related: math.stackexchange.com/questions/51752/… $\endgroup$ – Asaf Karagila Jan 9 '13 at 0:55

The empty set is indeed a set (the set of no elements) and it is a subset of every set, including itself. $$\forall A: \emptyset \subseteq A,\;\text{ including if}\;\; A =\emptyset: \;\emptyset \subseteq \emptyset$$

$$\text{BUT:}\quad\emptyset \notin \emptyset \;\text{ (since the empty set, by definition, has no elements!)}$$

That is, being a subset of a set is NOT the same as being an element of a set: $$\quad\subseteq\;\, \neq \;\,\in: \;\; (\emptyset \subseteq \emptyset), \;\;(\emptyset \notin \emptyset).$$

$\emptyset \;\subseteq \;\{1, 2, 3, 4, 5\},\quad$ whereas $\;\;\emptyset \;\notin \;\{1, 2, 3, 4, 5\},\;$.

$\{3\} \subseteq \{1, 2, 3, 4, 5\},\quad$ whereas $\;\;3 \nsubseteq \{1, 2, 3, 4, 5\}, \text{... but}\; 3 \in \{1, 2, 3, 4, 5\}$.

  • 1
    $\begingroup$ Yes that makes sense because $\emptyset = \emptyset$ $\endgroup$ – Leonardo Jan 8 '13 at 4:12
  • $\begingroup$ Leonardo, Indeed, you are right; do you see the difference between being an element of and being a subset of? $\endgroup$ – Namaste Jan 8 '13 at 4:17
  • $\begingroup$ I understand now, thank you. $\endgroup$ – Leonardo Jan 8 '13 at 4:27
  • 1
    $\begingroup$ Your very welcome. It takes a little getting used to the distinction! $\endgroup$ – Namaste Jan 8 '13 at 4:29

Being a subset and being an element are different. For every set $A$, $\varnothing\subseteq A$, but not necessarily $\varnothing\in A$. As an example, we can consider $$A=\left\{1,2,3,\{1,2\}\right\}\text{ and }B=\{1,2,3\}.$$ Then $\{1,2\}\in A$ AND $\{1,2\}\subseteq A$ whereas $\{1,2\}\subseteq B$ but $\{1,2\}\notin B$. Hopefully this clarifies that being an element and a being a subset are different things.

  • $\begingroup$ That makes a bit more sense too.. according to your set definitions $\emptyset \notin A$, but, $\emptyset \subseteq A$, and also I think $\emptyset \subset A$ $\endgroup$ – Leonardo Jan 8 '13 at 4:18
  • $\begingroup$ @Leonardo: If $\subset$ means proper containment, then $\emptyset\subset A$ is equivalent to $A\neq\emptyset$. I.e., $\emptyset\subset\emptyset$ is false with that convention, but that is the only exception. (You are probably talking about the explicit example $A$ given above, but I wanted to be explicit because $A$ is also quantified at the beginning of the answer.) $\endgroup$ – Jonas Meyer Jan 8 '13 at 4:21
  • $\begingroup$ That is only because the empty set is the empty set. But A is not the empty set. Yes the empty set is not a proper subset of itself, I agree. For it would need to have an element of itself which itself does not have, which it has none to begin with. $\endgroup$ – Leonardo Jan 8 '13 at 4:23
  • $\begingroup$ Suppose $A\subset B.$ Then $x\in A \to x\in B$, Now suppose we examine $A=B=\varnothing$. It is true that $\varnothing \subset \varnothing$, because $x\in \varnothing\to x\in \varnothing$ is true: because because both the antecedent and consequent are false, and $F\to F = T$. $\endgroup$ – Namaste Apr 26 '17 at 16:50

$\varnothing$ is a subset of $\varnothing$, but $\varnothing$ is a not an element of $\varnothing$, because $\varnothing$ has no elements by definition.

For every set $A$, there is no element of $\varnothing$ that is not in $A$, and therefore $\varnothing\subseteq A$. This is true whether or not $A$ is empty.

For an example of a nonempty set that doesn't have the empty set as an element, consider the set $A=\{\{1\}\}$. That is, $A$ is the set whose only element is the set $\{1\}$. Because $\{1\}$ contains the element $1$, it is not empty, $\{1\}\neq\varnothing$. Because the only element of $A$ is not $\varnothing$, $\varnothing\not\in A$. The set $\{\{\varnothing\}\}$ also does not have the empty set as an element for the same reason. On the other hand, the set $\{\varnothing\}$ does have the empty set as an element. By definition, $\varnothing$ is the only element of the set $\{\varnothing\}$.

  • 1
    $\begingroup$ Yes there is no element of $\emptyset$ that is not in $A$ because it has none. $\endgroup$ – Leonardo Jan 8 '13 at 4:20
  • $\begingroup$ @Leonardo: That's right, and the only way for a set $B$ to not be a subset of a set $A$ is for $B$ to have some element that is not in $A$. $\endgroup$ – Jonas Meyer Jan 8 '13 at 4:23

Sorry to clutter the site with yet another answer to the many good ones already posted, but I'd like to offer one that is a bit more formal.

The empty or null set, $\emptyset$, is defined as a set that has no elements. A set $A$ is said to be a subset of $B$ if and only if every element of $A$ is also an element of $B$, notation $A \subseteq B$, or stated using predicate logic, $$\forall x(x \in A \rightarrow x \in B)$$

So to claim that the null set is a subset of any other set asserts $$\emptyset \in \emptyset \rightarrow \emptyset \in A$$ The logical proof of this simply observes that the antecedent of the implication is false therefore the implication is vacuously true. It is false because the definition of the empty set is one that contains no elements and therefore it does not contain itself. Since we have satisfied the definition of subset, we have proven the claim that the empty set is a subset of every set regardless of how counter intuitive it may be.

Hey, that's logic!

  • 2
    $\begingroup$ "So to claim that the null set is a subset of any other set asserts $\emptyset \in \emptyset \rightarrow \emptyset \in A$" how on earth did you arrive to that conclusion??? $\endgroup$ – Asaf Karagila Apr 23 '15 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.