Argue that a countable set minus a scalar quantity is still countable My question is as follows:
Assume we have a set $A \subseteq$ $\mathbb R$ that is both bounded (bounded is the only word here, there is no information of being bounded from below or low) and countable. Can we argue that $A - c$ is also countable, where $c \in \mathbb Z$?
My following argument was the following: $A$ is bounded and countable (premise stated above). Subtracting $c$ from all $a \in A$ only shifts the bound of $A$ by $-c$ units, but does not change the notion that $A-c$ is bounded. Moreover, just like $A$, $A-c \subseteq \mathbb R$. Therefore, $A-c$ is countable.
 A: Your argument for boundedness is correct, but but more precision is always good. By the way: a subset of $\Bbb R$ is said to be bounded if it is both upper and lower bounded. My proof would be the following:

If $A$ is bounded there are some real constants $K$, $M$ such that $K\le x \le M$ for every $x\in A$. Then, for every $y\in A-c$ we have $K-c\le y\le M-c$.

Your argument for countability is wrong. You only say that $A-c\subset \Bbb R$ and this alone does not imply at all that $A-c$ is countable. In fact "most" of subsets of $\Bbb R$ are not countable. You have to use that $A$ is countable and to build a bijection from $A$ to $A-c$. Something like that:

Define $f:A\to A-c$ to be $f(x)=x-c$ for $x\in A$. The function is well defined since $x-c\in A-c$ for every $x\in A$. Moreover is injective because $f(x)=x-c=y-c=f(y)$ implies that $x=y$. Finally, $f$ is surjective because if $y\in A-c$ then $y+c\in A$ and then $f(y+c)=y$. Thus, $f$ is bijective and hence $A$ and $A-c$ have the same cardinality.

