Is there a mistake in this question: $\forall a\in A: |\{x\in A:x\le a \}|=|\{ y\in B :y\le a \}|$?

Question

Two ordered sets $(A,\le_A), (B,\le_B)$ and there's an isomorphic function $f:A\to B$

Prove $\forall a\in A: |\{x\in A:x\le a \}|=|\{ y\in B :y\le a \}|$

I think there's a mistake in this question, how can you compare elements of $A$ with elements of $B$ ? with which order do you compare them and what if they're disjoint sets ?

Shouldn't it be $|\{ y\in B :y\le f(a) \}|$ ?

Answer 1

Yes it should be f(a), and the isomorphism required to compare the cardinality is just f restricted to $\{x\epsilon A:x\leq a\}$

Answer 2

Yes, you are right, there should be $f(a)$ instead of $a$.