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i am reading a textbook here and i saw, there is notion of Complement of a function. or Negation of a function definiton, this is whow i understood but it is definitely wrong how i do it, i know. in the textbook i have a function $f$: $$f: \{2n | n\in \mathbb{N}_0 \}: n \rightarrow n+1$$

and they said, that its complement(Negation) is: $f': \{2n+1| n\in \mathbb{N}_0 \}$

how they are coming to this? Generally how do i find the complement of a function? isnot just the image of a given function?

or the problem underlying is: how do i find the injective extension of a function using this complement?

DEFINITION of how to find injective extension: in order to extend f with whole $\mathbb{N}$ to $F: \mathbb{N} \rightarrow \mathbb{N}$, one has to find the injective map $f'$ of complement of $\{2n | n\in \mathbb{N_0}$ in $\mathbb{N_0}$ to the complement of image of $f$.

thanks a lot for help

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Can you perhaps quote directly from the book, precisely as it is given in the book. Make sure to include the relevant definitions. – Ittay Weiss Feb 3 '13 at 5:54
@IttayWeiss, i think, i am confused. they took the complement of image set. and they call the complement of a function the function which is mapping to that complement image set. does it make sense? – doniyor Feb 3 '13 at 5:58
no, it does not make sense. That is why I asked for the precise definitions from the book. – Ittay Weiss Feb 3 '13 at 5:58
I can try it in German but can't make any guarantees – Ittay Weiss Feb 3 '13 at 6:01
sorry, that doesn't help. – Ittay Weiss Feb 3 '13 at 6:08
up vote 2 down vote accepted

I am working very specifically from this particular example. You have the function

$$f:\{2n:n\in\Bbb N_0\}\to\Bbb N_0:n\mapsto n+1\;.$$

This function is an injection from the subset $\{2n:n\in\Bbb N_0\}$ of $\Bbb N_0$, and the problem is to extend it to an injection $F:\Bbb N_0\to\Bbb N_0$.

The idea is then to define a new function $f\,'$ so that if we then let

$$F:\Bbb N_0\to\Bbb N_0:n\mapsto\begin{cases}f(n),&\text{if }n\in\operatorname{dom}f\\f\,'(n),&\text{if }n\in\operatorname{dom}f\,'\;,\end{cases}$$

this $F$ will be an injection whose domain is all of $\Bbb N_0$. This means that the domain of $f\,'$ must be the complement of the domain of $f$: we want

$$\operatorname{dom}f\,'=\Bbb N_0\setminus\operatorname{dom}f=\{2n+1:n\in\Bbb N_0\}\;,$$

the set of odd natural numbers. We also want $F$ to be injective, so we want to make sure that the range of $f\,'$ is disjoint from the range of $f$: we want $\operatorname{ran}f\,'\cap\operatorname{ran}f=\varnothing$, or, equivalently,

$$\operatorname{ran}f\,'\subseteq\Bbb N_0\setminus\operatorname{ran}f=\Bbb N_0\setminus\{2n+1:n\in\Bbb N_0\}=\{2n:n\in\Bbb N_0\}\;.$$

That is, the range of $f\,'$ should be a subset of the complement of the range of $f$. In this case we can actually make the range of $f\,'$ all of the complement of the range of $f$: there is an easy bijection

$$f\,':\{2n+1:n\in\Bbb N_0\}\to\{2n:n\in\Bbb N\}\;;$$

it is in fact the inverse of the function $f$.

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