Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Since every countably infinite set is the range of a bijective function defined on $N=\{1,2,3,\dots\}$, we may regard every countable set as the range of a sequence of distinct terms. Speaking more loosely, we may say that the elements of any countable set can be "arranged in a sequence."

Does the opposite statement make sense ? i.e. if we can arrange the elements of a set as the elements of a sequence of distinct elements (for example, if that set is an infinite subset of a countably infinite set, say $X$, then using the above argument for $X$) then that set is countably infinite. I am confused.

share|cite|improve this question
How do you define "arrange in a sequence?" Most rigorous definitions would be as the $1-1$ range of a function on $\mathbb N$. – Thomas Andrews Aug 14 '13 at 0:28

That's exactly what "arranging in a sequence" is. You essentially create a bijective map from a $\mathbb{N}$ to a set $A$. To see this, lets say we arrange the elements of $A$ into the sequence $a_1, a_2, a_3, \dotsc$ with each $a_i$ unique. Then by using the indices $1, 2, 3, \dotsc$, you've implicitly defined the map $i \mapsto a_i$. To make it explicit, just define the function $a: \mathbb{N} \to A$ by $a(i) = a_i$. It is easy to see it is bijective since $a_i$ are unique.

So yes you can go the other way. If you can arrange the elements into a sequence of distinct terms, then the set is countable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.