Is the set of all integers not divisible by 5 countable?

Having a hard time picturing this because it seems almost every set is countable. Why can’t every number in a set be matched up with the natural numbers?

How do I know/show this is countable?

Is 1,2,3,4,6,7,8,9,11,12,13,14,16…. count as a pattern or is that not a pattern?

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Hint: it is at most countable, and it is not finite.
– dxiv
2 days ago

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The fact that you’re asking if a sequence of numbers is a “pattern” in this context shows that there are some serious foundational issues with your understanding. Before we can effectively help you, you need to go and review the definitions of finite and countable. I am confident the word “pattern” will not appear in those definitions.
– Nesos
2 days ago

  

 

Yes. It is at least countably infinite since {2k:k∈N}\{2^k : k \in \mathbb{N}\} is a subset of it; at most because it’s a subset of N\mathbb{N}.
– MathematicsStudent1122
2 days ago

  

 

Most sets are uncountable. But all the sets of integers are countable.
– Asaf Karagila
2 days ago

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2 Answers
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Let Am={5k+m:k∈Z}A_m=\{5k+m:k \in \mathbb{Z}\}, 1≤m≤41 \leq m \leq 4. Then you set is just ∪41Am\cup_{1}^{4}A_m. Every AiA_i is countable and union of countables is countable.

Subset of countable sets are countable. Since the set of all integers not divisible by 55 is a subset of integers, which is clearly countable, it is countable.