Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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.)
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.
$A\subseteq \emptyset \iff \forall x\in A, x\in \emptyset$
$\not\exists x: x\in \emptyset$
$\therefore A\subseteq \emptyset \iff A=\emptyset$
