-2
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Zain Patel, Shailesh, C. Falcon, pjs36
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $f(m) = n$

$\endgroup$
  • $\begingroup$ 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

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