Intuition vs Proof of the empty set as a universal subset My Discrete Mathematics book talks about the proof of why null set is the subset of every non-empty set. It has raised some questions in my head.
Firstly, Why? I mean why will we want to make simple, intuitive things so complicated. I mean why is there need to say tht non-empty set contains an empty set? We can see that a non-empty set has elements and combination of atleast one of them will be a subset of the main set which makes perfect sense. But why will we need to bring up the whole counter intutive idea of null set being a subset? Same question about the set being the subset of itself. 
What I am looking for is that if somebody can help me understand the purpose and importance of doing so.
 A: What you write is not true at all. Take $X=\{\varnothing\}$, and look at $Y=\{X\}$. The set $Y$ has one element, which is $X$. So $Y$ has one element which is neither empty, nor a subset of $Y$.
On the other hand, $\varnothing\subseteq Y$, because every element of $\varnothing$ is an element of $Y$. Simply because $\varnothing$ has no elements!
There are three main reasons to prove that $\varnothing\subseteq A$ for any set $A$:


*

*It is an excellent opportunity to discuss "vacuous truth" or "vacuous implications", where something is true simply because it cannot be false (i.e., there are no counterexamples).

*It establishes that $\varnothing$ is the "smallest" possible set, and that it is unique (if $A$ and $B$ are two empty sets, then $A\subseteq B$ and $B\subseteq A$, so $A=B$).

*It is a true statement, with an easy and accessible proof, so presenting it is a very good idea. And it's an excellent opportunity to exercise the definition of what it means for one set to be a subset of another.
A: I wouldn't exactly say that it's important that $\varnothing$ is a subset of every set. Sure, in some contexts it probably is, but that's not really the point.
The point is that from the definition of "subset", we can prove that $\varnothing$ is a subset of all sets; this simply must be so, if we agree on the definitions. So, what's important here is understanding the definition and understanding how to draw conclusions from the definition;  using definitions to prove things.
That is what's important here, not necessarily the results themselves (in my opinion, in this specific situation).
A: The reason for these conventions is to make certain "abstract" statements true in more cases than if there weren't these definitions.
For instance, once can say that $n\leq m$ holds if any only if the set $\{1,\ldots,n\}$ is a subset of $\{1,\ldots,m\}$. There's nothing to argue about if $1\leq n,m$ and also $n\not=m$.
Now what about if $n=0$? Since $0\leq m$ for any natural number $m$, we would want the set $\{1,\ldots,0\}$ to be contained in $\{1,\ldots,m\}$. But what does $\{1,\ldots,0\}$ even mean? We could say that it is the set of all integers $k$ such that $1\leq k\leq 0$, namely the empty set. This would then force us to say that the empty set is a subset of all sets of the form $\{1,\ldots,m\}$, if we wanted our pattern to continue.
Also for the case $n=m$, if we wanted to the pattern to hold we have to say that every set is a subset of itself.
Long story short: We make these definitions so that patterns continue to hold in all cases.
A: It's due to axiom of specification which states that to every set A and every condition S(x) there exists a set B whose elements are exactly those elements of A for which S(x) is true.
Now if your condition is such that no element of x satisfies that condition,this will create an empty set. And if your condition is such that every element satisfies it, then B=A. 
from here A and empty set are considered as subsets.
A: $A$ is a subset of $B$ exactly when every element of $A$ is also an element of $B$.  Consider what it means, then, for $A$ to not be a subset of $B$;  it means precisely that there is some element of $A$ that is not an element of $B$.  
Since there are no elements of $\emptyset$, then for any set $C$, there cannot be something which is an element of $\emptyset$ and not an element of $C$, so $\emptyset$ must be a subset of $C$.
