proving $|X|<|Y|$, $|Y|<|Z| \Longrightarrow |X|<|Z|$ without CSB

Question

how to prove that if $|X|<|Y|$, $|Y|<|Z|$ then $|X|<|Z|$ without CSB theorem? it is immediate that $|X|\leq |Z|$ so I tried to assume that $|X|=|Z|$ and reach a contradiction but so far I couldn't.

Answer 1

I think is a matter of definitions.

Let $≤$ be a preorder. We define the induced equivalence by $x \sim y \iff x ≤ y$ and $x ≥ y$. We also define the induced strict variant by $x < y \iff x ≤ y$ and $x \nsim y \iff x ≤ y$ and $x \ngeq y$.

Now we can prove the following general theorem: If $≤$ is a preorder, then $<$ is a strict order (i.e. an antireflexive transitive relation).

In your case, the preorder is $|X| ≤ |Y|$ defined as “there is an injection from $X$ to $Y$”. There is also an equivalence $|X| = |Y|$ defined as “there is a bijection between $X$ and $Y$”. This equivalence is finer than the equivalence induced by the preorder. However, CSB exactly says that they are the same. So if you define $|X| < |Y|$ as the strict variant, you get the result from the general theorem, but if you defined it using the finer equivalence, you need CSB.

In fact, it can be proved that you need CSB. You can prove CSB from your proposition if the definition is based on the finer equivalence. Let $|X| ≤ |Y|$ and $|Y| ≤ |X|$. If there is no bijection between $X$ and $Y$, then $|X| < |Y| < |X|$ so $|X| < |X|$, which is a contradition.