# prove whether functions are injective, surjective or bijective

Hello,

I have a few questions regarding the above mentioned task. One has to show whether or not these functions are injective, surjective or bijective.

This seems straightforward for most of these functions:

(1) Not injective since some values are hit multiple times. Also not surjective since not every y has a corresponding x.

(2) Bijective, since there is a one to one correspondance.

(3) Bijective, since it's only from the natural numbers, so every value will be hit exactly once

My question now is: How do I properly phrase this? Some of these seem quite intuitive but I'm struggling with the correct notation of this.

• $(4)$ is surjective as $\{n\}\in \mathcal P(\Bbb N)\backslash\{\emptyset\}$ for each $n\in \Bbb N$. But $(4)$ is not injective as $\min\{1,2\}=1=\min\{1,3\}$. – Mathlover Sep 22 '20 at 8:35
• You should use the definitions of the stated concepts. I.e. if you want to show that a function is not injective then provide two points $x_1\neq x_2$ such that $f(x_1)=f(x_2)$. If you want to prove the contrary, then show that $f(x_1)=f(x_2)$ implies that $x_1=x_2$. – Maximini Sep 22 '20 at 8:39
• For (4), it is important to know whether $0\in\Bbb N$. This is not universally agreed on one way or the other, so it should be clarified. – Arthur Sep 22 '20 at 8:39
• okay I'll try to do that thank you! 0 is an element of the natural numbers in this course. I was not sure if it's enough if I just take 2 points and calculate them to show that there exists an injection or not – 23408924 Sep 22 '20 at 8:41
• @0-thUser thanks a lot!. I'm never quite sure if this is "enough" for a proof or if i need to provide more details – 23408924 Sep 22 '20 at 8:48

$$\bullet (1)$$ takes $$-1$$ and $$0$$ to the same place, so it isn't injective. I doubt it's surjective, either, because we need a solution to $$y=x^2+x$$ for any $$y$$. Take $$y=-1$$, say. $$x^2+x+1$$ has no real roots, since the discriminant is negative. Complete the square and you get $$y=(x+1/2)^2-1/4$$. Thus you can graph and "see" that nothing less than $$-1/4$$ is hit. It's a parabola (opening upwards), after all.

$$\bullet(3)$$ isn't surjective: not every natural is a fourth power

$$\bullet (4)$$ is of course not injective, but surjective: two different sets can have the same $$\inf$$; there is a set with any natural as $$\inf$$

• oh you're right. I completely forgot. So (3) can only be injective. – 23408924 Sep 22 '20 at 8:40
• $(2)$ looks good. there is an inverse. – Chris Custer Sep 22 '20 at 8:44
• Another way of putting it in the first one is that a parabola doesn't pass the horizontal line test, so is not injective. And has a max or min, so is not surjective. – Chris Custer Sep 22 '20 at 9:21
• Thanks! yeah I mean this seems intuitive, however i mostly struggle with the notation. I'm not sure if they accept it without enough details though. But i think i should be able to write it down correctly now – 23408924 Sep 22 '20 at 9:23
• Great! The right amount of rigor can be an issue... Reminds me of the time my linear algebra professor drew waving hands on the board (to indicate hand waving argument). – Chris Custer Sep 22 '20 at 9:25

(4) It is not injective, since many subsets may share the same smallest element. But it is surjective, since for any natural number $$n$$ there is a larger one $$n+1$$ such that both lie in the some subset in $$\mathcal{P}(\mathbb N)$$.