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Problem: Show that the set of odd numbers is countable

Attempt: So for this problem, I just need to find a bijection from the natural numbers to the set of odd numbers. However, I find the claim "odd numbers" a bit ambiguous because it can be odd natural numbers or odd integers. However, I think that problem is about the odd integers. Would it be sufficient to show a bijection in the following manner?

1->1
2->-1
3->3
4->-3
5->5
6->-5

and so on.

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    Consider $2n - 1$ for all $n \in \mathbb{N}$. – 2017-01-12
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    The set $A$ of odd integers is a subset of the integers, which are countable. So $A$ is at most countable. – 2017-01-12
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    You edited your question, so I deleted my answer. Your bijection is correct. – 2017-01-12
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    @TheMathNoob Still interested an explicit function of the bijection you made? – 2017-01-12
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    Yes, it would be nice to see it. – 2017-01-12
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    @ΘΣΦGenSan how would you go about showing the function below is a bijection? – 2017-09-22
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    @Michelle Drolet Just show that the mapping is 1-1 and onto. – 2017-09-23
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    @ΘΣΦGenSan I know that I have to do that, I am just not sure how to show the onto portion. – 2017-09-23
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    @ Michelle Drolet The definition is canonical. In fact, $$f(\mathbb{N})= \{\dots,-5,-3,-1,1,3,5,\dots\}.$$ – 2017-09-23

2 Answers 2

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Your answer is correct but in case you want an explicit function then the following help . Define $f:\mathbb{N}\to \{\dots,-5,-3,-1,1,3,5,\dots\}$ by

$$ f(n)= \begin{cases} n&, \mbox{if $n$ is odd}\\ 1-n&, \mbox{if $n$ is even} \end{cases} $$ for all $n\in\mathbb{N}$.

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Let $X$ be the set of odd positive integers. Then if we can create a bijection between $X$ and $\mathbb{N}$, then we are done. Why? Because a bijection exists, the number of elements in $X$ and $\mathbb{N}$ must be the same.

So let $f:\mathbb{N}\to X$ be $f(n)= 2n-1$, which is clearly a bijection.

Then you can say the odd integers are $X\cup Y$ where $Y=\{-1\cdot x~\mid~x\in X\}$. Then countable unions (in this case a finite union) of countable sets are countable.

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    how does your function map negative integers? – 2017-01-12
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    I think that somehow it has to be a piece-wise function – 2017-01-12
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    Check my edited answer. – 2017-01-12
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    Nice, I think that now it makes sense. However, how can you make sure that the changes that you effected on the set X to create the set Y don't affect it's count-ability?. I hope I spelled that right. What is the noun of countable?. – 2017-01-12
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    I suppose you can establish a bijection between $X$ and $Y$ with $g:X\to Y$ as $g(x)=-x$. Since $X$ is countable, then $Y$ is countable. It is a fairly obvious observation however. – 2017-01-12
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    That's great. Thank you. – 2017-01-12