Trivial question about set difference: does $B\subseteq A$ imply $B\setminus\{x\}\subseteq A\setminus\{x\}$? Suppose we have two sets $A$ and $B$ with $B \subseteq A$ does it follow that 
$$ B \setminus \{ x \} \subseteq A \setminus \{ x \} $$
I cant find a counter example so i assume this is true in general?
 A: Suppose $y\in B\setminus\{x\}$. Then $y\in B$ and, since $B\subseteq A$, we have $y\in A$. Further, $y\in B\setminus\{x\}$ implies that $y\neq x$, and so $y\in A\setminus\{x\}$. Therefore, $B\setminus\{x\}\subseteq A\setminus\{x\}$. 
A: If $b\in B\setminus \lbrace x\rbrace$, then $b\in B$ and $b\neq x$. Since $B$ is a subset of $A$, then we know $b\in A$. So $b\in A$ and $b\neq x$, by definition it follows that $b\in A\setminus \lbrace x\rbrace$.
A: The statement $B\subseteq A$ means every member of $B$ is a member of $A$. The question is then whether every member of $B\smallsetminus\{x\}$ is a member of $A\smallsetminus\{x\}.$
Suppose $y$ is a member of $B\smallsetminus\{x\}.$ Then $y$ is a member of $B$ and $y\ne x.$ Since $y$ is a member of $B$ and every member of $B$ is a member of $A$, it follows that $y$ is a member of $A.$ And it is still true that $y\ne x.$
Thus every member of $B\smallsetminus\{x\}$ is a member of $A\smallsetminus\{x\}.$
A: for any two sets $A,B$ the relation $B \subseteq A$ is equivalent to $B \cap A=B$. let $X$ be the complement of $\{x\}$ in $A \cup B \cup \{x\}$, so $A\setminus\{x\} = A \cap X$, and $B\setminus\{x\} = B \cap X$. we have
$$
(A\setminus\{x\})  \cap (B\setminus\{x\}) = ( A \cap X) \cap (B \cap X) = (A \cap B) \cap X = B \cap X = B\setminus\{x\} 
$$
hence
$$
 B\setminus\{x\} \subseteq  A\setminus\{x\} 
$$
