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

(4) Don't quite understand where to start with this one.

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.

  • 1
    $\begingroup$ $(4)$ is surjective as $\{n\}\in \mathcal P(\Bbb N)\backslash\{\emptyset\}$ for each $n\in \Bbb N$. But $(4)$ is not injective as $\min\{1,2\}=1=\min\{1,3\}$. $\endgroup$ – Mathlover Sep 22 '20 at 8:35
  • $\begingroup$ 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 $x_1=x_2$. $\endgroup$ – Maximini Sep 22 '20 at 8:39
  • $\begingroup$ 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. $\endgroup$ – Arthur Sep 22 '20 at 8:39
  • $\begingroup$ 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 $\endgroup$ – 23408924 Sep 22 '20 at 8:41
  • $\begingroup$ @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 $\endgroup$ – 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 $\inf$

  • $\begingroup$ oh you're right. I completely forgot. So (3) can only be injective. $\endgroup$ – 23408924 Sep 22 '20 at 8:40
  • $\begingroup$ $(2)$ looks good. there is an inverse. $\endgroup$ – Chris Custer Sep 22 '20 at 8:44
  • $\begingroup$ 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. $\endgroup$ – Chris Custer Sep 22 '20 at 9:21
  • $\begingroup$ 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 $\endgroup$ – 23408924 Sep 22 '20 at 9:23
  • $\begingroup$ 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). $\endgroup$ – 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 $\mathcal{P}(\mathbb N)$.


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