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I'm trying to follow the proof of the Cantor-Bernstein Theorem in Jech's Set Theory: The Third Millenium Edition.

If $\lvert A\rvert \leq \lvert B\rvert$ and $\lvert B\rvert \leq \lvert A\rvert$, then $\lvert A\rvert =\lvert B\rvert$

The beginning of the proof is:

If $f_1:A\to B$ and $f_2:B\to A$ are one-to-one, then, if we let $B'=f_2(B)$ and $A_1=f_2(f_1(A))$, we have $A_1\subset B'\subset A$ and $\lvert A_1\rvert=\lvert A\rvert$.

And I think I understand this part well, but the very next conclusion:

Thus we may assume that $A_1\subset B\subset A$ and that $f$ is a one-to-one fuction of $A_1$.

I'm lost as to why we can now assume that B is a subset of A. In fact, I don't see why A and B might not be totally disjoint. How do the facts of the first sentence of the proof allow us to assume that?

Pat Muchmore
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  • Because we can replace $B$ with $B'$, there is a bijection between them. – Asaf Karagila Oct 05 '24 at 13:11
  • I was assuming that $f_1$ and $f_2$ were one-to-one into A and B respectively, but not full bijections, so I guess that's part of my big problem. But, although I understand that the bijection between $B$ and $B'$ means they have the same cardinality, how can we assume that the sets themselves are the same? $B'$ is made of elements from $A$, not $B$. – Pat Muchmore Oct 05 '24 at 14:37

1 Answers1

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This is a standard reduction of the problem to a special case. Here is the structure of the proof.

Lemma: Suppose $A_1\subseteq B\subseteq A$ and $|A_1|=|A|$. Then $|A|=|B|$.

(This is what Jech proves.)

Theorem: If $|A|\leq |B|$ and $|B|\leq |A|$, then $|A|=|B|$.

Proof: Let $f_1\colon A\to B$ and $f_2\colon B\to A$ be one-to-one. Let $B'=f_2(B)$ and $A'=f_2(f_1(A))$. Note that $f_2\colon B\to B'$ is one-to-one and onto $B'$, so $|B|=|B'|$. And $f_2\circ f_1\colon A\to A'$ is one-to-one and onto $A'$, so $|A|=|A'|$.

Now $A'\subseteq B'\subseteq A$ and $|A|=|A'|$. Apply the Lemma with $B=B'$ and $A_1=A'$. The Lemma says $|B'|=|A|$, and since $|B|=|B'|$, we have $|B|=|A|$. $\square$

Note that the $B$ in the Lemma is not the same $B$ as in the statement of the Theorem! We take the $B$ in the theorem, produce a new $B'$, and apply the Lemma to that $B'$.

Now, frequently when the statement of the lemma is a special case of the statement of the theorem, and the full theorem reduces to this special case, an author will write "we may assume (we are in the special case)" rather than separating out the special case as a lemma as I did above.

For example, when proving that every non-zero integer $n$ has a unique factorization into primes, a proof may begin "multiplying by $-1$ if necessary, we may assume $n>0$", and then it may use, for example, that if $m\mid n$, then $m\leq n$ (which is not true for negative $n$). When we then try apply the proof to factor $-40$, we do not think the author actually meant to assert that $-40>0$! Instead, the proof instructs us to factor $40$, and then obtain a factorization of $-40$ from the factorization of $40$.

Alex Kruckman
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  • Thank you for this. I feel like I'm missing something really stupid here though: I'm fine with the idea that $B$ and $B'$ have the same cardinality, but doesn't the second part of the proof imply that A and B share specific elements? It seems like $A_1$, $A$, and $B'$ are all made up of elements of $A$. – Pat Muchmore Oct 05 '24 at 14:42
  • @PatMuchmore Yes, $B'\subseteq A$. This is the point of the reduction, to reduce to this easier case. We're replacing $B$ (which may be disjoint from $A$) with another set $B'$ of the same cardinality which is a subset of $A$. Now we show $|B'|=|A|$, from which it follows that $|B|=|A|$, since $|B|=|B'|$. – Alex Kruckman Oct 05 '24 at 14:47
  • RIght, but the second line specifically says $A_1\subset B\subset A$. Doesn't that imply that they aren't disjoint, or do you think there's a misprint here? – Pat Muchmore Oct 05 '24 at 15:05
  • @PatMuchmore No, there's no misprint, and Jech's proof is correct as written. I've edited my answer to spell out more explicitly how the reduction to the special case works. I think it's really a notational thing that's tripping you up, namely the replacement of $B$ by $B'$ in the argument. Advanced textbooks like Jech's will assume the reader has the requisite mathematical maturity to follow arguments like this. – Alex Kruckman Oct 05 '24 at 19:57
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    @Pat Muchmore: Advanced textbooks like Jech's will assume --- Given that Jech has also written an undergraduate level text, you might find it helpful to have a copy of it to look at for more detailed exposition of the "basic stuff". Or any similar level text (such as some of those listed here; but note that other books listed there are at or near the advanced-Jech text level), but Jech's elementary text is likely to better match what you see in Jech's more advanced text. – Dave L. Renfro Oct 05 '24 at 20:15
  • @DaveL.Renfro Thanks to both of you. You are, of course, right that I don't have the requisite mathematical maturity, but I've been having trouble finding a text suitably "in between" the Enderton Set Theory text that I feel like I have a pretty good handle on and this one. I really want to gain a stronger understanding, in particular, of large cardinal axioms and this Jech text was at least more comprehensible to me than the Kanamori. I'll see if the more basic Jech is better for me due to notation matching up with this advanced one. – Pat Muchmore Oct 05 '24 at 22:10
  • @PatMuchmore I highly recommend the book by Hrbacek and Jech. One very nice thing about it is that while it's suitable for an undergraduate course, it includes a survey of more advanced topics in set theory in the later chapters (I'm referring to the 3rd edition). In particular, Chapter 13 is an introduction to large cardinal properties. – Alex Kruckman Oct 05 '24 at 23:05
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    @Pat Muchmore: In this MSE answer I listed several "next level" set theory texts, and some of them might be of interest to you, such as Drake/Singh [2], Hajnal/Hamburger [4], Just/Weese [7] [8], Levy [12]. – Dave L. Renfro Oct 06 '24 at 00:18