1

Is it possible for there to not exist surjection from sets $X$ to $Y$ when there is also no surjection from $Y$ to $X$? I am strongly included to say no, as the former seems to imply that $|X| < |Y|$ whereas the latter implies the opposite- however, we made assumptions that there exists a mapping which is "one-to-one" enough and doesn't go to the same thing every time when concluding this (which seems a fair assumption, but which I can't formally justify) and if $X$ and $Y$ were infinite sets, perhaps these rules break down.

Princess Mia
  • 2,403
  • If $A \subset B$ (a propery one) then crearly there isnt a surjection from $A$ to $B$. On the other hand, we can define $f: B \rightarrow A$ as $f(x) = x$ if $x \in A$ and $f(x)=a$ for a constant element $a \in A$.

    More simple, $f$ is the identity when restringed to $A$ and in map $B \backslash A$ to some element $a \in A$.

     – Brian Britos Simmari Sep 26 '23 at 19:25
  • If you work in ZFC then the answer is: no, because any two sets are comparable in cardinality and you cannot have $|X|<|Y|<|X|$. I don't know what the situation is in ZF – FShrike Sep 26 '23 at 19:35
  • @BrianBritosSimmari first of all: of course there can be a surjection $A\to B$ even if $A\subset B$. You will always find such $A$ when $B$ is infinite. In fact it can be treated as definition of infinity. More importantly: OP is not assuming $X\subset Y$ or vice versa. – freakish Sep 26 '23 at 20:09

1 Answers1

2

The answer to that question depends on the kind of set theory you are working with. The most common nowadays is ZF (and its variations), which stands for Zermelo-Fraenkel set theory.

Now, in this set theory, it indeed is possible to have two sets without surjection between them. Yes, counterintuitive.

But it is also typical to consider ZFC, which is the ZF set theory together with the Axiom of Choice. This axiom changes everything, in some sense the theory is a lot more restrictive. The following statements are provable within ZFC:

  1. For any two sets $X$ and $Y$ at least one of the following is true: $|X|\leq |Y|$ or $|Y|\leq |X|$. Or in other words: any two sets are comparable, which is also known as the Principle of Cardinal Comparability. By the way, it is equivalent to the Axiom of Choice.
  2. $|X|\leq |Y|$ if and only if there is surjection $Y\to X$. This statement is known as the Principle of Partition and you can find the proof here. As far as I understand it is still an open question whether the Principle of Partition implies the Axiom of Choice, see this.

which together answer your question.

I am strongly included to say no, as the former seems to imply that $|X| < |Y|$

Only under the Axiom of Choice. So as you can see, the devil is in the details...

freakish
  • 42,851