8

If $A$ and $B$ are finite sets it is true that $|A|\leq |B|$ iff there is an injection from $A$ to $B$ and $|A|\geq |B|$ iff there is an surjection from $A$ to $B$. This motivates me to define these inequalities for arbitrary sets.

But doesn't this definition makes the Cantor-Schroeder-Bernstein theorem trivial?

What is wrong here? I should have to prove the Cantor-Schroeder-Bernstein theorem before defining these inequalities for arbitrary sets? Or they don't make sense at all?

Later
  • 722
  • 2
  • 8
  • 24
Gabriel
  • 4,513
  • 2
    Of course they make sense. The CBS theorem is what tells you that, if $|A|\le|B|$ and $|B|\le|A|,$ then $|A|=|B|.$ This is nontrivial for infinite sets. – bof Dec 07 '16 at 10:32
  • To appreciate that, as @bof said, this is in fact nontrivial for infinite sets, just go ahead and try to prove it. You will likely run into a situation where it seems you just have too little information to conclude that there is in fact a bijection $A \cong B$. Without a hint, CBS is pretty tricky to prove. However, if you assume the axiom of choice, it's basically trivial to verify. So the key point is to prove it in $\operatorname{ZF}$. – Stefan Mesken Dec 07 '16 at 10:47

2 Answers2

20

If you define the relation $\le$ on arbitrary sets by $$ |A|\le |B| :\iff \text{ there is an injection }A \to B $$ then you need the Cantor-Schröder-Bernstein theorem to prove (that is something you have to prove) that $\le$ is an order relation. Just because you call something $\le$, it does not mean that $$ |A|\le |B|, \ |B|\le |A| \implies |A|= |B| $$ holds true. That is Cantor-Schröder-Bernstein.

martini
  • 84,101
  • 3
    $|\bullet|\le|\bullet|$ is not an order anyway. Your point is of course that it induces a partial order on the isomorphisms classes in the category of sets. – Carsten S Dec 07 '16 at 13:29
  • @CarstenS Partial? – Jack M Dec 07 '16 at 18:17
  • 1
    @Jack: Yes. Linear orders are partial orders. – Asaf Karagila Dec 07 '16 at 18:18
  • @CarstenS Thanks, of course. – martini Dec 08 '16 at 07:42
  • @JackM, I assumed that we are working in ZF.because the fun part of Cantor-Schröder-Bernstein is that it is a theorem in ZF. And without AC two sets need not be comparable. http://math.stackexchange.com/questions/151904/does-the-assertion-that-every-two-cardinalities-are-comparable-imply-the-axiom-o – Carsten S Dec 08 '16 at 13:54
5

First, let me point out that if $B$ is empty and $A$ is not empty, then $|A|\geq|B|$, but there are no surjections from $A$ onto $B$.

Secondly, for infinite sets, it's entirely unclear why $|A|\leq|B|$ and $|B|\leq|A|$ should imply $|A|=|B|$. It's easily enough when you can actually write down the functions by hand. But if I had just told that there is an injective function from $\Bbb Q$ into $\Bbb N$, how would you propose to turn it into a bijection "off the cuff"?

Sure, in the case of $\Bbb N$ it's relatively easy. So let's press on. How would you suggest to construct a bijection between $\Bbb Q$ and the algebraic numbers? Or between $\ell_5$ and $\ell_2$?

As the comments point out, the main difficulty is to prove the theorem without using the axiom of choice. Cantor somewhat dismissed it as trivial once the axiom of choice is given, or rather the well-ordering theorem. And indeed this is almost trivial if you know that every set has a minimal ordinal with which it can be bijected and the basic properties of ordinal comparability.

But without the axiom of choice? Without the axiom of choice things become much more difficult. And in fact, the statement "$|A|\geq|B|$ if there is a surjection from $A$ onto $B$ [or $B$ is empty]" is no longer true, because there might not be an injection from $B$ into $A$, despite there is a surjection from $A$ onto $B$.

The triviality is somewhat of a cultural consequence. You've been born—mathematically speaking—into a world where the theorem is well-established; possibly you're a second, third or even fourth generation of people who learned mathematics when the theorem was already an established result of mathematics. This makes you far more susceptible to taking it for granted. Just as a kid today if they can imagine a world without televisions, or if they are young enough, a world without smartphones, the internet, Google. For them it's as trivial as the Cantor–Bernstein theorem is to you.

Asaf Karagila
  • 393,674
  • Excellent answer! About the first statement ($|A|\geq |B|$ iff there is a surjection from $A$ to $B$): is it true if $B$ is not empty? If so, why? – Gabriel Dec 07 '16 at 18:45
  • 1
    Yes, assuming the axiom of choice, this is true. And for finite sets this is of course true in general. The reason is that if $f\colon A\to B$ is surjective, then for each $b\in B$, the set $A_b={a\in A\mid f(a)=b}$ is non-empty, so we can choose from it and get an injection from $B$ into $A$. If $B$ is finite, then the axiom of choice is not needed to prove that such a choice function exists, as ZF proves that for the finite case. – Asaf Karagila Dec 07 '16 at 18:50
  • Seems great! Thank you! – Gabriel Dec 07 '16 at 18:51