If $A$ is a set with $$ elements and $C ⊆ A$ is a set with $1$ element, then $\setminus C$ is a set with $−1$ elements Asked in a homework to prove this how does this look?
Let $A=\{1,2,3,4,...,m-1,m\}$.
Since $C⊆A$,let $C={x}$ where $x\in A$ and $1\leq x \leq m$, let $x=m$.
Hence, $A\setminus C$=$\{1,2,3,4,....,m-1\}$
Therefore, $A\setminus C$ will have $m-1$ elements.
 A: I think you've got the right idea, but to make the proof a little nicer I think you should make it more general. For instance, let $A=\{x_1,\ldots,x_m\}$ so that $A$ is just a set with $m$ elements, i.e. the elements are not necessarily the numbers $1,\ldots,m$. Then, since $C\subseteq A$ has one element, we can write $C=\{x_j\}$ for some $1\leq j\leq m$. Now you shouldn't say "let $x=m$", you should just leave it as some $x_j$. Then $A\setminus C=\{x_1\ldots,x_{j-1},x_{j+1},\ldots,x_m\}$ has $m-(j+1)+1+(j-1)=m-1$ elements.
A: These statements are tricky because of their very intuitive nature: one does not know what to assume and what to prove. Note that $A$ has $m$ elements, but that does not mean that necessarily $A = \{1, \dots, m\}$. The set could be 'anything', as long as it has $m$ distinct elements. 
You should start by asking yourself what 'having $m$ elements' means, or rather, how your lecturer defined it. An excessively formal but correct definition could be that you have a bijection $\sigma : \{1, \dots, m\} \to A$. If these are terms you are unfamiliar with, this is just a fancy way of saying that you can enumerate the elements of $A$, 
$$
A = \{ a_1, \dots, a_m\}.
$$
Since $C \subset A$, the elements of $C$ are elements of $A$. Now, since $C = \{x\}$ has a single element and $x \in A$, there exists some $1 \leq i \leq m$ for which $x = a_i$. Otherwise, $x \neq a_i$ for all $i$ and thus we would actually have that $x \not  \in \{a_1, \dots, a_m\} = A$. Now, by definition once again, 
$$
A \setminus C = \{y \in A : y \not  \in C\} = \{y \in A : y \neq a_i\} = \{a_j : 1\leq j \leq m, j \neq i \}.
$$
It remains to prove that the latter has $m-1$ elements. Depending on the formality you want, you could just say that this set is exactly
$$
\{a_1, \dots, a_{i-1},a_{i+1},\dots,a_m\}
$$
and handwave the possible index problems (what if $i = 1$?). A surely correct way is to use the previous definition, and prove that 
$$
\begin{align}
\sigma :  \{1,\dots, m-1& \}  \to \{a_j : 1\leq j \leq m, j \neq i \} \\
& j \longmapsto \cases{a_j \text{ if $j < i$ } \\
a_{j+1} \text{ if $j \geq i$ } \\}
\end{align}
$$
is a bijection. 
