1

As title says, I have some infinite set of integers $A$, a function $f:A \to \mathbb N$ defined by $f(n) = |\{n' \in A| n' < n\}|$ is surjective.

I'm having problems proving it. I'm not entirely sure it is correct either.

Suppose $A = \mathbb Z$. Then if for example I have $f(n) = 0$, then that means that there are no integers smaller than $n$. Which seems strange. Hence there is no $n \in \mathbb Z$ such that $f(n) = 0$ and so $f$ is not surjective.

Where is my mistake? Question taken from Asaf Karagila's answer here Proof that aleph null is the smallest transfinite number?

Community
  • 1
Oria Gruber
  • 12,739
  • $A$ should be an infinite subset of $\mathbb{N}$ for this to be correct (in particular, your counter-example does not work). – Clément Guérin May 15 '16 at 16:11
  • Although Asaf wrote integers in his answer to the other question, he clearly meant non-negative integers. More generally, the result is true for all $A\subseteq\Bbb Z$ that have a least element. – Brian M. Scott May 15 '16 at 16:27
  • As @Brian points out, the meaning was $\Bbb N$ and not $\Bbb Z$. It is common, at least in set theory, to refer to the natural numbers as "the integers". Unfortunate terminological differences are unfortunate indeed. – Asaf Karagila Jul 07 '16 at 12:38

1 Answers1

0

I think that the function is not well defined. Your example that $A = \mathbb{Z}$ and $n = 0$ gives the value $f(n) = \mathbb{Z}\setminus \mathbb{N}$. This give us a problem, because your codomain is $\mathbb{N}$. If your function is not well defined, it can't be a function and therefore can not be surjective. If you change the codomain to $\mathbb{R}$ it should be true I believe.

Mariëtte Schonfeld
  • 16