I'm really confused on how to go about proving this . I had to skip class because of my uni transfer orientation ( go UTD! ) and I'm having a hard time understanding this concept.


closed as off-topic by Zain Patel, Shailesh, C. Falcon, user8795, pjs36 May 15 '17 at 1:24

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  • 3
    $\begingroup$ Try the map $\;n\to3n\;$ from the naturals to your set. $\endgroup$ – DonAntonio Mar 16 '16 at 21:53
  • 1
    $\begingroup$ What is your definition of countable? $\endgroup$ – hardmath Mar 16 '16 at 21:54
  • 2
    $\begingroup$ Any subset of a countable set is countable. Your set is a subset of the naturals. $\endgroup$ – Ross Millikan Mar 16 '16 at 21:54
  • $\begingroup$ Ok so heres what i got. " Set {3, 6, 9, 12 , 15...} is a subset of N since its an injection of N. the set is countable and a subset of a countable set is also countable. thus the set {3 , 6 , 9 , 12 , 15 ... } is countable as well since its a subset of N." does it look right? $\endgroup$ – shayan javadi Mar 16 '16 at 22:04

A comment sayst that your set: $$S = \{3, 6, 9, 12, 15, ...\} \subset \mathbb{N}$$ because every number in your set is an element of $\mathbb{N}$. Because all natural numbers are countable, your set $S$ is countable.


Perhaps your definition of countable is actually what some call "countably infinite" in which case it is not true that any subset of a countable set is countable. With this definition $\{0\} \subseteq \mathbb{N}$ is not countable. So definitions are important in this question as one commenter points out.

If your definition of countable is "countably infinite" then to show a set $A$ is countable, we must find a bijection between $A$ and $\mathbb{N}$. Another commenter gives you the function to use i.e. consider $$f : \mathbb{N} \to \{3,6,9,\ldots\}$$ where $$f(n) = 3n$$

It is not hard to show this function is a bijection, i.e. injective and surjective. To do this show:

injectivity: $n \neq m \implies f(n) \neq f(m)$

surjectivity: $n \in \{3,6,9,\ldots \} \implies $ there is $m \in \mathbb{N}$ such that $f(m) = n$

  • $\begingroup$ Hey, what does $f: \mathbb{N} ...$ mean? $\endgroup$ – Obinna Nwakwue Mar 28 '16 at 23:48
  • $\begingroup$ Sorry for the late response, it means $f$ is a function with domain $\mathbb{N}$ and codomain $\{3,6,\ldots\}$ $\endgroup$ – Ryan Sullivant Apr 15 '16 at 4:05

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