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I am working on these problems:

Prove:

(a) If $|A|<|B|$ and $|B|\leq|C|$, then $|A|<|C|$.

(b) If $|A|\leq|B|$ and $|B|<|C|$, then $|A|<|C|$.

As far as I know, it is clear that $|A|\leq|C|$ in both cases, so I just want to show that $|A|\neq|C|$, for which I was trying to prove by contradiction.

If I assume that $|A|=|C|$, then the first one will yield: $(|A|<|B|)\land(|B|\leq|A|)$.

Similarly, the second one will yield: $(|A|\leq|B|)\land(|B|<|A|)$.

Yes, I know that in such case $|A|\neq|B|$, so both of them reduce to $(|A|<|B|)\land(|B|<|A|)$.

Well, I have no idea how to prove that $|A|<|B|$ and $|B|<|A|$ cannot hold simultaneously.

Of course, I don't think we can prove it by so-called "transitivity", because in order to show that $(|A|<|B|)\land(|B|<|C|)\implies|A|<|C|$, we still need to show that $|A|\neq|C|$, which will reduce to exactly the same problem.

So I hope if anyone had some good ideas on this proof. Any help will be appreciated.

P.S. It is not a homework problem. I am reading Introduction to Set Theory by K. Hrbacek and T. Jech to learn some ideas about axiomatic set theory by myself. This problem is from section 4.1.

Update: It is just something like the law of trichotomy of cardinal numbers. But I don't want to use transfinite induction or AC. I wonder if there is some elementary approach to this proof.

Update: In the book, $|A|=|B|$ if there is a bijection from $A$ to $B$. $|A|\leq|B|$ if there is an injection from $A$ to $B$. In particular, if $|A|\leq|B|$ and $|A|\neq|B|$ (the contropositive of $|A|=|B|$), then $|A|<|B|$.

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  • It may help to prove that $\leqslant$ is an equivalence relation on the cardinality of sets. – Math1000 Dec 18 '19 at 04:29
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    Umm. I don't think $\leq$ is an equivalence relation on the cardinality of sets. It is, more or less, like a partial order. –  Dec 18 '19 at 04:32
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    The Cantor-Bernstein Theorem shows that if you have an injection $f\colon A\to B$ and an injection $g\colon B\to A$, then there is a bijection $g\colon A\to B$. In particular, if you had $|A|\lt|B|$ and $|B|\lt|A|$, you would get a bijection $A\to B$, contradicting that $|A|\neq|B|$ (which is implicit in “$|A|\lt|B|$”). – Arturo Magidin Dec 18 '19 at 04:43
  • @BernardPan Yes, that is what I meant to say. Thank you for correcting me. – Math1000 Dec 18 '19 at 04:44
  • What cardinal order axioms are you using and/or whose treatment are you following? (There are a few distinct starting points...) – Eric Towers Dec 18 '19 at 04:46
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    @ArturoMagidin That's it. Thank you. Before I was thinking that it was just a corollary from the transitivity of $\leq$ on the cardinality of sets, no need to apply the Cantor-Bernstein theorem. (I haven't even looked at the proof of it yet, but I'll do it right now) –  Dec 18 '19 at 04:56
  • @EricTowers Not yet. I am just at the start point of this chapter. I have no idea how the cardinal order is formalized in this book. But after all, thank you for your response. –  Dec 18 '19 at 05:00
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    Can you please state the exact definitions of $\le$ and $<$ as used in the book? – Nate Eldredge Dec 18 '19 at 07:03
  • @NateEldredge Definitions updated. –  Dec 18 '19 at 19:35

1 Answers1

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Well, I have no idea how to prove that $|A|\lt|B|$ and $|B|\lt|A|$ cannot hold simultaneously.

As far as I know, the standard definition of $|A|\lt|B|$ is "there is an injection from $A$ to $B$, but there is no injection from $B$ to $A$". (Unfortunately, your book defines $|A|\lt|B|$ as "there is an injection from $A$ to $B$ but there is no bijection between them." Fortunately, in view of the Cantor–Bernstein theorem, that is equivalent to the natural definition, which is the one I gave.)

If $|A|\lt|B|$ and $|B|\lt|A|$ hold simultaneously, then we have: $$\text{there is an injection from }A\text{ to }B;\tag1$$ $$\text{there is no injection from }B\text{ to }A;\tag2$$ $$\text{there is an injection from }B\text{ to }A;\tag3$$ $$\text{there is no injection from }A\text{ to }B;\tag4$$

All you need is basic logic to get a contradiction from $(1)$ and $(4)$, or from $(2)$ and $(3)$.

If $|A|\lt|B|$ and $|B|\le|C|$, than $|A|\lt|C|$.

From $|A|\lt|B|$ and $|B|\le|C|$, we have injections $A\to B\to C$; composing these we get an injection $A\to C$. We also have to show that there is no injection $C\to A$. If we had an injection $C\to A$, composing this with the given injection $B\to C$ would provide an injection $B\to A$, contradicting the assumption that $|A|\lt|B|$.

bof
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  • The definition of $<$ is $\le$ but $\ne$, that is, $|A<|B|$ iff there is an injection from $A$ to $B$ but there is no bijection. – Andrés E. Caicedo Dec 18 '19 at 12:51
  • An unfortunate choice, since it means we beed the nontrivial Cantor–Bernstein theorem to show that it's equivalent to the natural definition. – bof Dec 18 '19 at 12:56
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    I think the natural definition is the one I mentioned. We are just too used to the equivalence. As the comments beneath the question indicate, Cantor-Schroeder-Bernstein was indeed the missing point. – Andrés E. Caicedo Dec 18 '19 at 12:58
  • I think the natural definition is the one that would be preferred in the more general situation, where they are not equivalent. If $\le$ is a quasi-ordering (reflexive transitive relation, sometimes called a "pre-ordering"), $a\lt b$ naturally means that $a\le b$ and $b\not\le a$; the relation "$a\le b$ and $a\ne b$" is not particularly useful as far as I know. That's how I'd define $a\lt b$ for general order types; isn't it standard? – bof Dec 18 '19 at 13:10
  • Would you say that $\eta\lt1+\eta$ and $1+\eta\lt\eta$? (Where $\eta$ is the order type of the rational numbers.) – bof Dec 18 '19 at 13:14
  • I defer to your expertise. – Andrés E. Caicedo Dec 18 '19 at 13:20
  • @bof Thanks a lot for your answer. I am also wondering that how can we deduce from "there is an injection from $A$ to $B$" and "there is no bijection from $A$ to $B$" to "there is no injection from $B$ to $A$". What if there is an injection from $B$ to $A$? It still reduces to the statement "if there is both an injection from $A$ to $B$ and an injection from $B$ to $A$, then there is a bijection between $A$ and $B$", which is exactly the one we want to prove. (or is proven by CSB theorem) –  Dec 18 '19 at 19:28
  • @BernardPan Yes, that equivalence is precisely the CSB theorem. Is it the point of this exercise to prove CSB, or has CSB already been established earlier in the book? If you are asking about how to prove the CSB theorem, that has been asked and answered several times on this site. You can find a concise proof here. – bof Dec 18 '19 at 22:46
  • @AndrésE.Caicedo I believe "I defer to your expertise" translates to "I won't waste my time arguing with this fool." :-) – bof Dec 18 '19 at 22:50
  • I defer to your expertise on the subtleties of language as well. :-) – Andrés E. Caicedo Dec 18 '19 at 22:52
  • @bof Actually CSB has been proven at the second part of that section. But this problem appears at the beginning of the exercise list. I didn't expect it would be concerned with CSB so I tried solving it before I dug into the proof of CSB. After all, thank you for your response. –  Dec 19 '19 at 03:02