# Understanding cardinality notation

The title isn't completely clear, but let me try to explain. The notation $$|A| \leq |B|$$, even for infinite sets, implies that there exists an injection $$A \hookrightarrow B$$. If I write $$|A| < |B|$$ thereafter, it means there is no surjection $$A \to B$$, so we can say, informally, that $$B$$ is "larger" in some sense than $$A$$.

Suppose I've established, via a chain of maps, $$|A| = |B| < |C| = |D| \leq |E| = |F|.$$ I want to then say that $$|A| < |F|$$, which I would be able to do using the usual rules of arithmetic. Surely I can inject $$A$$ into $$F$$: just compose injections. But how do I know there is no surjection from $$A$$ to $$F$$? There must a way to find a contradiction, in particular, one of these equalities I know about would disappear if I assumed the existence of a surjection, in which case $$|A| = |F|$$, but I can't find it after multiple attempts.

One possibility is contradiction. Assume instead $$|A| \geq |F|$$, i.e., there exists an injection $$F \hookrightarrow A$$. Then $$|F| \leq |A| = |B| < |C| = |D| \leq |E| = |F|,$$ but this doesn't really tell me anything, because I can't read off $$|F| < |F|$$ without using the same "transitive" property I'm trying to prove above.

Because there is an injection $$C\to F,$$ and $$C$$ isn’t empty, there is a surjection $$F\to C.$$
So if there is a surjection $$A\to F,$$ there is a surjection $$A\to C.$$
Lemma: If $$f:X\to Y$$ is an injection, and $$X$$ is non-empty, there is a surjection $$g:Y\to X.$$
Proof: Let $$x_0\in X$$ be any element. Define: $$g(y)=\begin{cases}f^{-1}(y) &\text{when }y\in f(X)\\ x_0&\text{otherwise} \end{cases}$$
• Yes. $|A|=|B|$ means that, since there is no surjection $B\to C,$ there is no surjection $A\to C.$ May 26, 2021 at 15:14