I am trying to understand whether the set is finite, countable, or uncountable.

$$\{x \in\Bbb Q \mid 1<x<2 \} \qquad\text{is countable. }$$

but i dont understand why though. is it countable because there is finite numbers between 1 and 2? how could you count all the numbers between 1 and 2? why cant it be uncountable??

An extra problem i was givien: $$\left\{ \frac mn \mid m,n \in \Bbb N, m<100, 5<n<105\right\}$$ is finite. i think it is finite because $\Bbb N$ is greater than or equal to 1. N > 1. i would like to know if im wrong or right. I want to understand everything i do.

Thank you for efforts and time


  • $\begingroup$ I have no idea what the sets you have given us are supposed to be. Could you please explain with words what the sets are supposed to be? $\endgroup$ – Andrew D Jun 5 '13 at 16:41
  • $\begingroup$ im so sorry. i meant to say {x is belong to Rational number l 1 is less than x is less than 2} $\endgroup$ – amie Jun 5 '13 at 16:44
  • 1
    $\begingroup$ @AndrewD Amie is using 'l' as a vertical bar; apparently his or her terminal has a sans-serif font. $\endgroup$ – MJD Jun 5 '13 at 16:48
  • $\begingroup$ @amie: Here's how to enter mathematical formulas on this site: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – MJD Jun 5 '13 at 16:50
  • $\begingroup$ @Amie: Do you know the proof that the set $\Bbb Q$ of all rational numbers is countable? Do you remember what a countable set is? $\endgroup$ – MJD Jun 5 '13 at 16:56

Countable means you can put it in bijection with $\Bbb N$ (some people include finite sets as countable, but you seem to mean countably infinite). Have you seen the proof that all of $\Bbb Q$ or $\Bbb {N \times N}$ is countable? Your first is a subset of this, so is clearly countable. There are an infinite number of rationals between $1$ and $2$.

For the second, there are only $100$ choices for $m$ (or $99$ if you do not include $0$) and $99$ choices for $n$, and $100 \times 99$ is clearly finite.


There are several reasons for why the first set given is countable; the way I would prefer to do this is by the result that a countable union of countable sets is countable:

If we think about the rationals in their usual form of $\frac{p}{q}$ for some $p,q \in \Bbb N$, then for $q=2$ we have countably many choices of $p$ such that our rational number is between $1$ and $2$. As we can then repeat this process for $q \in \Bbb N $ such that $q>1$, we see that each time we have countably many choices for $p$, and hence as we have a countable union of the sets (for a given $q$)

$$ \{p \in\Bbb N \mid 1<\frac{p}{q}<2 \} $$

it follows that your original set must be countable (and as our union is over a countably infinite set, it follows that your set must also be countably infinite, and cannot be just finite). You can then use a similar style of proof to prove that the whole of $\Bbb Q$ is countable.


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