An exercise of "Universal Algebra: Fundamentals and Selected Topics" of Clifford Bergman.
(a) Let $P$ be a poset. Show that there is no surjective, isotone map from $P$ to $<Dn(P), \subseteq>$. (Hint: suppose $f$ were such a map. Consider the set $B = \{x \in P : \exists(y)\ x\le y \ \ \& \ y \notin f(y) \}$.)
(b) Use part (a) to prove Cantor’s theorem: for any set X, there is no surjective function from X to Sb(X).
Here $Dn(P)$ is the set of downsets of $P$.
(b) follows from (a) if we see $X$ as a poset with the trivial partial order $x \le y$ iff $x=y$. With this order relation $Sb(X)=Dn(X)$ and every map from $X$ to $Sb(X)$ is isotone, hence there is no surjective map.
It is intuitive, making some attempts, that (a) is true at least for finite sets. I'm interested in a proof that uses the definition of $ B $, which I can't even prove to be non-empty.
The set B is nonempty.
Reason: since f is surjective, something gets mapped to the empty set. If f(x) = ∅, then x ∈ B since ∃y (e.g., y = x) with the given properties.
Next observe that B is a downset, so f maps something in P to B.
Let b ∈ P be such that f(b) = B.
If b ∉ B, then there is no y ≥ b such that y ∉ f(y).
But what about b itself? Indeed, we have b ≥ b and b ∉ B = f(b), so b satisfies the criteria for membership in B (contradiction).
Since assuming b ∉ B leads to a contradiction, assume b ∈ B.
Then there exists y ≥ b such that y ∉ f(y). Let f(y) = C (so y ∉ C).
Now use monotonicity to obtain the contradiction y ∈ C.