We have $A = \mathbb{R}$ and $B = \{x|x \in \mathbb{R} \land \exists y (y\in \mathbb{Z} \land |x-y| < \frac{1}{2})\}$

How is $A-B$ countably infinite?

I know the definition of set minus is $A \cap \overline{B}$, but I don't know how to translate $B$ and to see if this is a countable set.

  • 1
    $\begingroup$ Consider - what real numbers are not in $B$? Argue that it is those that are precisely between the integers, i.e. numbers of the form $n + 1/2$ for $n \in \Bbb Z$. That set is countable. $\endgroup$ – Eevee Trainer 2 days ago
  • 4
    $\begingroup$ don't we have $A-B=\{\ldots, -2.5, -1.5, -0.5, 0.5, 1.5, 2.5, \ldots\}$? $\endgroup$ – angryavian 2 days ago
  • $\begingroup$ How do you get all of this multiples of halves? I thought when $x-y < 1/2$ right? So, how do we get a half? $\endgroup$ – DippyDog 2 days ago
  • $\begingroup$ $|x-y|<1/2\iff x\in(y-1/2,y+1/2)$ and thus $B=\{x\in\Bbb R|\exists y\in\Bbb Z(x\in(y-1/2,y+1/2))\}$. Now letting $\{y\}$ denote the fractional part of $y\in\Bbb R,\{x\}<1/2\iff\lfloor x\rfloor\le x<\lfloor x\rfloor+1/2$ and $\{x\}>1/2\iff\lceil x\rceil-1/2<x<\lceil x\rceil$. Thus the only real $x$ which do not belong to $B$ are those which have $\{x\}=1/2$ i.e. $A-B=\{z+1/2:z\in\Bbb Z\}$. $\endgroup$ – Shubham Johri 2 days ago
  • $\begingroup$ I'm not sure how to say this in formal terms, but a rough argument can go like this: The real numbers x not in B must make $|x-y| \geq 1/2$ for any integer $y$. But if $|x-y|$ is strictly greater than 1/2, there exists some integer $y'$ that makes $|x-y'| < 1/2$. So the complement of B contains only real numbers $x$ such that $|x - y| = 1/2$ for any integer $y$ $\endgroup$ – c1620 2 days ago


$$A-B=\{z+\frac{1}{2}\mid z \in \Bbb Z\}$$ so it is in bijection with $\Bbb Z$.

If $x$ is a real and its fractional part is $< \frac12$ then $\lfloor x \rfloor$ witnesses that $x \in B$, so $x \notin A-B$. If the fractional part is $>\frac12$ then $\lceil x \rceil$ witnesses that $x \in B$ and also $x \notin A-B$ so the only reals in this difference are those with exactly fractional part $\frac12$, i.e. the set on the right hand side.


$|x-y|<1/2\iff x\in(y-1/2,y+1/2)$ and thus $B$ contains all intervals of the form $(y-1/2,y+1/2),y\in\Bbb Z$. The only real numbers not in $B$ are of the form $y+1/2,y\in\Bbb Z$ and thus $A-B=\{y+1/2:y\in\Bbb Z\}$ is countably infinite since $\Bbb Z$ is countably infinite.


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