How does one prove this? Can one prove by contradiction by saying:

Let $A$ be any set such that $A$ contains at least one element. Now assume $A \subset \emptyset$. This is clearly absurd by the definition of $\emptyset$, so $A$ is not a subset of the empty set, and the only subset of $\emptyset$ is $A$ such that $A = \emptyset$.

Does this reasoning make sense? Help is much appreciated, this is my first time writing proofs.

  • 2
    $\begingroup$ *$\subseteq{{}}$ $\endgroup$ – Git Gud Jul 17 '14 at 22:42
  • 5
    $\begingroup$ Hint: $\emptyset\subseteq A$ holds for all sets $A$. $\endgroup$ – Berci Jul 17 '14 at 22:42
  • $\begingroup$ It is more or less ok, but it seems to me that you only "proved" the $\Rightarrow$ implication. (The other proposed Berci) $\endgroup$ – Peter Franek Jul 17 '14 at 22:46
  • $\begingroup$ math.stackexchange.com/questions/330676/… $\endgroup$ – mle Jul 17 '14 at 23:20

The definition of $\varnothing$ refers to its lack of elements, it says nothing about its subsets, so although it is clearly absurd that a nonempty set should be a subset of $\varnothing$, it's not enough to say "it is clearly absurd"... at least, not for a proof of such a low-level statement.

Something along the following lines would be better: suppose $A \ne \varnothing$. Then there exists $x \in A$. But since $A \subseteq \varnothing$, it follows that $x \in \varnothing$, which is absurd by definition of $\varnothing$.

Notice that this proof differs from yours in that the contradiction ('absurdity') follows from statements about elements instead of about subsets.

(Of course, a proof that doesn't use contradiction follows from the easy-to-prove fact that $\varnothing \subseteq A$ and by definition of set equality.)

  • $\begingroup$ I find the prejudice against the statement unsettling. $\endgroup$ – Git Gud Jul 17 '14 at 22:48
  • $\begingroup$ I gave your answer the points because it was the most thorough and points out the issues in my proof. Although, what does Git Gud mean by his comment? What is the prejudice against the statement? $\endgroup$ – mathjacks Jul 20 '14 at 22:01
  • $\begingroup$ @flapjackery: He was making a joke because I called it a "low-level" statement (in this sense rather than this sense of course :P) $\endgroup$ – Clive Newstead Jul 20 '14 at 22:03

For all sets $A$, $\emptyset \subseteq A$, so by the definition of set equality,

$$ A \subseteq \emptyset \implies A = \emptyset $$

And the converse is clear, that is if $$ A = \emptyset $$ then $$ A \subseteq \emptyset $$

(also try and refrain from using $\subset$ if set equality is possible, it can be confusing)


This is a similar question to being asked to prove that $\emptyset \subseteq A$ for any set $A$. To prove it, you need to show if $x \in \emptyset$, then $x \in A$. But the statement $p \implies q$ is true even when the statement $p$ is false, because $False \implies True$ is a true statement. And for any $x$, $x \in \emptyset$ is a false statement. (Note that $False \implies False$ is also a true statement, so it does not matter if $x \in A$.)

Having said that, to prove $A \iff B$, you need to prove $A \implies B$ and $A \impliedby B$.

Let's prove the $\implies$ direction first: suppose $A \subseteq \emptyset$. We want to prove $A = \emptyset$. Suppose by contradiction that $A \neq \emptyset$. Since $\emptyset \subseteq A$, then there must exist $x \in A$ such that $x \not \in \emptyset$. But if $x \in A$ and $x \not \in \emptyset$, then $A \not \subseteq \emptyset$, which contradicts our assumption.

Now for the $\impliedby$ direction: suppose $A = \emptyset$. Then since any set is a subset of itself, we have that $A \subseteq \emptyset$, as desired.


$A\subseteq \emptyset \iff \forall x\in A, x\in \emptyset$

$\not\exists x: x\in \emptyset$

$\therefore A\subseteq \emptyset \iff A=\emptyset$


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