Is the null set a subset of every set? Ever since day one of of my Mathematical Logic course, this fact has really bothered me. I cannot wrap my head around how an empty set is a subset of every possible set. Could someone kindly explain how this is true? Any help is appreciated! 
 A: A remarkably simple way of trying to grasp why that is the case is as follows:

We can consider any set and throw away all its elements, then we are
  left with the subset { }. This means that { } is a subset of any set.

A: If you're comfortable with proof by contrapositive, then you may prefer to prove that for any set $A,$ if $x\notin A,$ then $x\notin\emptyset$. But of course, $x\notin\emptyset$ is trivial since $\emptyset$ has no elements at all. Hence, $x\notin A\implies x\notin\emptyset,$ so by contrapositive, $x\in\emptyset\implies x\in A,$ meaning $\emptyset\subseteq A$.
A: By definition, $A$ is a subset of $B$ if every element of $A$ is in $B$.
If we set $A=\emptyset$, then the above statement is vacuously true. Every element of $A$ is in fact an element of $B$ since the former has no elements.
A: I'm practicing my set proving skills, so shoot me down if I'm wrong.
I'm going to use proof contradiction as an alternative approach since that approach has not been mentioned as answer in this topic.
So we want to prove:
${ \emptyset \subseteq A }$
In other words
${\forall_x [(x \in \emptyset) \to (x \in A)]}$ def. subset
So since proof by contradiction is used, lets try to prove the opposite and hope to run into a contradiction.
${\lnot \forall_x [(x \in \emptyset) \to (x \in A)]}$
After moving the negation inside, we get
${\exists_x[(x \in \emptyset) \land \lnot(x \in A)]}$
So we want to demonstrate there exists an element of the empty set that isn't in A.
since ${x \in \emptyset}$ will always be false we get
${\exists_x[F \land \lnot(x \in A)]}$
${\exists_x[F]}$ domination law
So there doesn't exist such an element. And this contradicts with our assumption that such an element exists. Therefor we have proven ${ \emptyset \subseteq A }$
A: I am quite comfortable with the other two answers but there is a softer way to answer this question. I think the following can be made formal using the Axiom of Choice (perhaps in conjunction with a function $Y\rightarrow \{0,1\}$), but if you really want to be that formal just use one of the two perfectly good arguments above.
This soft approach is as follows. Where $X$ and $Y$ are sets, a subset of $X\subseteq Y$ corresponds to a choice of elements from $Y$.
For example, where $Y=\{1,2,3,4\}$, a subset $X$ is formed by answering the questions
$$1\in X?,\,2\in X?,\,3\in X?,\,4\in X?$$
For example, the subset $\{1,4\}\subset Y$ corresponds to choosing one and four but not two and three.
If we choose all of the elements of $Y$ we have the full subset and so, as it consists of a choice of elements of $Y$ --- namely all of them --- we have that
$$Y\subseteq Y.$$
Now what if we choose none of the elements of $Y$? This is a choice of elements and so is a subset of $Y$. It is of course the empty set and in this sense we have
$$\emptyset\subset Y.$$
So a subset corresponds to a choice of elements from $Y$, choosing none is a choice, and therefore the empty set is always a subset.
You will need some of the more formal arguments from above to understand why $\emptyset\subseteq\emptyset$.
A: Below, an attempt at showing as intuitively as possible why one is compelled to say that the empty set is a subset of any set. 


