Let $A = \{1, 2, 3\}$.

Is the following expression true then?

$\varnothing \in A$.

I'm having issue understanding if the empty set is an element of $A$. Would love a brief explanation of why it is or isn't.

Thanks a lot.

  • 1
    $\begingroup$ The things between the braces are the elements of $A$. Answering the question "Is $4\in A$?" is precisely the same difficulty as "Is $\emptyset\in A$?" $\endgroup$ – rschwieb Feb 4 '14 at 19:53

$A$ is a set with three elements, namely $1,2$, and $3$. Since $\emptyset$ is none of these three, $\emptyset\notin A$.

It is indeed true that $\emptyset \subseteq A$. This means that each element of $\emptyset$ is also an element of $A$. Since $\emptyset$ has no elements, this statement is vacuously true, i.e. $\emptyset\subseteq S$ for all sets $S$. In fact even $\emptyset\subseteq\emptyset$.


No it is not an element of $A$. It is a subset of $A$, which we denoted $\varnothing \subset A$.


I'm not positively certain someone proved $\emptyset \subseteq X$:

Let there be a set $X$ such that $\emptyset \nsubseteq X$. So there exist $x \in \emptyset$ such that $x \notin X$. Wait, what?
QED ;-)

  • $\begingroup$ I find this line of reasoning to be rather convincing to students who are uneasy with the usual appeal to the vacuous conditional $x \in \emptyset \Rightarrow x \in X$. $\endgroup$ – Austin Mohr Feb 4 '14 at 21:14
  • $\begingroup$ @AustinMohr I'm not sure if your comment was a positive or negative remark..? Furthermore, It might be rationally/intelectually vacuous, but I'm not sure It's mathematically vacuous. What do you think? Furthermore, I actually had this as an exercise in chapter $0$ of Linear-Algebra I... :-) $\endgroup$ – user76568 Feb 4 '14 at 21:24
  • $\begingroup$ The statement "$x \in \emptyset \Rightarrow x \in X$" is vacuously true in a precise sense. Since there are no elements in the set $\emptyset$, the hypothesis "$x \in \emptyset$" is always false, and therefore the conditional as a whole is always true. (By the way, my remark about your answer was intended to be favorable.) $\endgroup$ – Austin Mohr Feb 4 '14 at 21:51
  • $\begingroup$ @AustinMohr I agree that the formal (And a tad un-intuitive) definition of implication should be acquired prior to elements of naive set theory. It is not the case though in formal education, at least in my country. $\endgroup$ – user76568 Feb 4 '14 at 22:04

You must take care to the definitions.

For every set $A = \{ a_1, a_2, ...a_n, ... \}$ its elements are the "things" between the braces.

The emptyset is defined as $\forall x (x \notin \emptyset)$, and is denoted also as $\{ \}$, in order to show that it has no elements.

The definition of $A \subseteq B$ is

$$\forall x (x \in A \rightarrow x \in B)$$


"every object that is an element of $A$ is also an element of $B$".

Now, if you apply this condition to $\emptyset \subseteq B$, by properties of conditional ($\rightarrow$) you may check that it is always satisfied, for every $B$, because of the above definition of $\emptyset$ : no $x$ is in $\emptyset$ and so ($P \rightarrow Q$ is true when $P$ is false) $\emptyset$ is a subset of every set.

But $x \in B$ is a different relation from $A \subseteq B$.

In your example, $1$ and $2$ and $3$ are the elements of $A$; so we have that $B = \{ 1,2 \}$ is a subset of $A$.

But $\emptyset$ is not "listed" between the braces, so it is not an element of $A$.

Of course, being a subset of every set, we have $\emptyset \subseteq A$.


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