Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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 $f$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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