As the question asked, Is every subset of $\mathbb{Z^+}$ countable?
Is every subset of $\mathbb{Z^+}$ countable?
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0Do you bother searching the site before asking? – 2012-03-15
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0Do you know the recursion principle? In my class: A infinite subset of $\Bbb Z_+$ is countably infinite was proved by exibhiting an explicit bijection with $\Bbb Z_+$. We had to check that the bijection map was well-defined and this motivated the recursion principle. – 2012-03-15
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0i couldn't search anything – 2012-03-15
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0Asaf this is a friendly environment, particularly after asking a question like that you should not post an answer to this question – 2012-03-15
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3@Kirthi: Please don't tell me what to do or not to do, what I should or should not do. I have my own moral compass and my own mind for making decisions like that. Thank you very much and have a pleasant time of the day. – 2012-03-15
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2Sorry I didn't mean to offend – 2012-03-15
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3@Asaf: We each have a moral compass. Maybe Kirthi is just expressing what his says in this situation... – 2012-03-15
2 Answers
Definition: $A$ is countable if and only if there exists a function $f\colon A\to\mathbb Z^+$ which is injective.
Now take the identity map, which is obviously injective since $id(x)=x$.
Definition: $A$ is countable if and only if there exists a function $g\colon\mathbb Z^+\to A$ which is surjective or $A=\varnothing$.
Now let $A\subseteq\mathbb Z^+$, if $A$ is empty then we are done. Otherwise pick $a\in A$ and define $g$ as follows:
$$g(n)=\begin{cases} n &n\in A\\ a &n\notin A\end{cases}$$
It is clear that this is a surjective function, as wanted.
Definition: $A$ is countable if and only if it is finite or there is $h\colon\mathbb Z^+\to A$ which is a bijection.
Suppose $A$ is a subset of $\mathbb Z^+$. If it is finite then we are done. Otherwise define by induction:
Now we claim that $h$ is a bijection. By induction we can easily show that $h(n) It is surjective since every $a\in A$ has only finitely many numbers smaller than itself therefore after at most $k$ steps (for some $k$) we have that $a Theorem: All three definitions above are equivalent. Proof: If $A$ is finite then this is clear. Suppose $A$ is infinite, if there exists a bijection of $A$ with $\mathbb Z^+$ then it is injective and its inverse is surjective, so the third definition implies the other two. If there exists a surjective function $g\colon\mathbb Z^+\to A$ define $f(a)=\min\{n\in\mathbb Z^+\mid g(n)=a\}$, since $g$ is a surjecitve function this is well-defined, and it is injective since $g$ is a function. Lastly if there exists an injection $f$ from $A$ into $\mathbb Z^+$ we can define $h\colon\mathbb Z^+\to\{f(a)\mid a\in A\}$ as in the third definition. Since $A$ is infinite the set to which we want $h$ to be defined into is infinite. Now $(f^{-1}\circ h)\colon\mathbb Z^+\to A$ is a bijection.
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0From your describe of the third definition, since A is a subset of $\mathbb{Z^+} $ what if$h(n)=max{a\in A}$ then how can h map n+1 to A as there are only finite number of element in A – 2012-03-15
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0@Mathematics: No, in the third option note that if $A$ is finite then we are done. **Otherwise** we define that function, namely $A$ is *not* finite, i.e. it is infinite. – 2012-03-15
You asked if for all sets $A \subset \mathbb{Z}^{+}$, is $A$ countable. Consider the set
$$A = \{1,2,3\} \subset \mathbb{Z}^{+}.$$
This set is a finite subset of $\mathbb{Z}^{+}$ that is not countable. Or in your question did you mean to ask if for all subsets of $\mathbb{Z}^{+}$ is it the case that they are at most countable?
Note: I am taking the definition of countable from Rudin's Principles of Mathematical Analysis that a set $A$ is said to be countable if there is a bijective function $f$ from $A$ to $\mathbb{Z}^{+}$.
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0That is an example *of* a countable subset. (Finite sets are countable too...) – 2012-03-15
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0@AsafKaragila I think OP was asking for at most countable.... – 2012-03-15
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2This is really about how you define countable. Finite sets are countable. – 2012-03-15
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0@AsafKaragila I have been told the definition given in Baby Rudin is not standard.... – 2012-03-15
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0The moral of the story is this: Never let analysts tell you about set theory. – 2012-03-15
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0This is why I was suggesting yesterday that there should be a way to communicate privately on a single question. – 2012-03-15
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3There seems to be no consensus whether "countable" allows finite set. To avoid misunderstandings it is a good idea always to write "at most countable" or "countably infinite", according to which sense one wants. – 2012-03-15
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0@HenningMakholm I edited my answer to reflect that. – 2012-03-15
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2I have never understood this lack of consensus. A set is countable if you can count it. Clearly you can count a finite set, while if there exists a bijection $f: \mathbb{N}\rightarrow S$ then $S$ is countable as $f(1)$ is the first element, $f(2)$ is the second element, ..., $f(n)$ is the $n^{th}$ element, etc. If finite sets are not allowed then the word "countable" was a bad choice! – 2012-03-15