Either two sets have the same cardinality, or one has cardinality greater than the other

This is a problem (10.11) from Munkres, Topology, 2 ed.

Problem: Let $A$ and $B$ be two sets. Using the well-ordering theorem, prove that either they have the same cardinality, or one has cardinality greater than the other.

Well-ordering theorem: If $A$ is a set, there exists an order relation on $A$ that is a well-ordering.

Two sets $A$ and $B$ are said to have the same cardinality if there is a bijection of $A$ with $B$. Let $A$ and $B$ be two nonempty sets. If there is an injection of $B$ into $A$, but no injection of $A$ into $B$, we say that $A$ has greater cardinality than $B$.

The hint to the problem makes reference to this theorem: Let $J$ and $C$ be well-ordered sets; assume that there is no surjective function mapping a section of $J$ onto $C$. Then there exists a unique function $h: J\to C$ satisfying the equation $$h(x)=\text{smallest}[C-h(S_x)].$$

I do not quite see the relationship between this last theorem to the problem at hand. The theorem implies that if there is no surjection of $A$ onto $B$ then there has to be an injection of $A$ into $B$. But how should I use it on the problem at hand? Thank you very much!

• If you know Zorn's lemma, you do not need to use the well-ordering theorem. – Andrés E. Caicedo Jan 9 '14 at 23:44
• math.stackexchange.com/questions/421638/… and a list of other threads: math.stackexchange.com/questions/421638/… – Asaf Karagila Jan 10 '14 at 10:38
• Interestingly, the statement you prove is actually equivalent to the well-ordering theorem: suppose any two cardinalities are comparable. Let $A$ be a set you want to well-order, and let $\aleph$ denote the Hartogs ordinal of $A$. By the definition of the Hartogs ordinal, there is no injection $\aleph\hookrightarrow A$, so, by hypothesis, there is an injection $A\hookrightarrow\aleph$. Now the well order on $\aleph$ induces a well order on $A$. – John Gowers Jan 13 '17 at 15:38

You say:

The theorem implies that if there is no surjection of $A$ onto $B$ then there has to be an injection of $A$ into $B$.

Then if there is no injection $A\to B$, then then there is a surjection $A\to B$ and hence an injection $B\to A$ (AC again). So $\lvert A\rvert\le \lvert B\rvert$ or $\lvert B\rvert\le \lvert A\rvert$.

By the way, the construction to define the function $h\colon J\to C$ actually either yields an order isomorphism from $J$ to an initial segment of $C$ or, if it fails, an order isomorphism from an initial segment of $J$ to $C$. In the latter case, this directly yields an injection from $C$ to $J$.

• Thank you for the answer. From your answer, I understand that if there is no injection of $A$ into $B$, then there has to be an injection of $B$ into $A$. In that case, $A$ would have greater cardinality than $B$. There is still one last thing I am not sure about: What if there is an injection of $A$ into $B$, as well as an injection of $B$ into $A$? Intuitively, I think that implies that there is a bijection of $A$ with $B$, but how should one demonstrate that? Thank you! – Fang Jing Jan 10 '14 at 21:19
• You are right, this is the Cantor-Bernstein Theorem. It's proof is elementary, but a bit tricky. – Carsten S Jan 11 '14 at 1:42

We may assume $A,B$ well-ordered. If there exists a surjection $g\colon A\to B$, then we obtain an injection $f\colon B\to A$, $b\mapsto \min g^{-1}(\{b\})$. If on the other hand there is no such surjection, then there is an injection $h\colon A\to B$. At any rate, $A$ and $B$ can be compared carinalitywise.

• You do not really need the well ordering for the first part. If $g\colon A\to B$ is a surjection then by the vanilla axiom of choice, $g$ has a right inverse $f\colon B\to A$. But now $f$ has a left inverse ($g$), so it is an injection. – John Gowers Jan 13 '17 at 15:40