Prove $A \subset \emptyset \iff A = \emptyset$ 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.
 A: 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.)
A: 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)
A: 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: $A\subseteq \emptyset \iff \forall x\in A, x\in \emptyset$
$\not\exists x: x\in \emptyset$
$\therefore A\subseteq \emptyset \iff A=\emptyset$
