How can I prove that $M\neq N \Leftrightarrow N\setminus M \neq \varnothing$ where $M\subset N$. I am doing some exam preparation assingment. I have came across this, and I dont know how to do it, so I was hoping someby can help me solve it. 
I need to prove that $M\neq N \Leftrightarrow N\setminus M \neq \varnothing$ where $M\subset N$.
Thanks
 A: HINT: Recall that $M\neq N$ if and only if $M\nsubseteq N$ or $N\nsubseteq M$.
A: If $M\subset N$ then $M=N$ if and only if they have the same elements.
Therefore $M\neq N$ if and only if there is an element of one which is not in the other. Since all elements of $M$ are in $N$ this implies that there is an element of $N$ which is not in $M$.
A: Assume that $M \subset N$, i.e. if $x \in M$ then $x \in N$. We have $M=N$ if, and only if, for all $x \in N$ we have $x \in M$. Hence $M \neq N$ if, and only if, there exists an $x \in N$ for which $x \notin M$. If there exists an $x \in N$ with $x \notin M$ then $x \in N\backslash M$ and hence $N\backslash M \neq \emptyset$
A: $(\Rightarrow)$ Suppose $M \neq N$ and $M \subset N$. Since $M \subset N$, every element of $M$ is also an element in $N$, that is, $a \in M \rightarrow a \in N$. However, since $M \neq N$, we can find an element that is in one set but not the other set. BUT, we know that every element in $M$ is also in $N$, so this means that we must be able to find an element in $N$ that is not in $M$. Therefore....
$(\Leftarrow)$ Suppose $M \subset N$ and $N \setminus M \neq \emptyset$. Since $M \subset N$, every element in $M$ is also in $N$, that is, $a \in M \rightarrow a \in N$. Now, since $N \setminus M \neq \emptyset$, we can find an element that is in $N$, but not in $M$. So, $N$ cannot be a subset of $M$ (if it was, every element of $N$ would be an element in $M$).
So, we have $M \subset N$ and $N \not \subset M$ which means .....
