This might be quite a naïve question, but I seem to be unable to find a good answer.
The set-valued functional $$ F(X) = (X \to X) \cup \{ \bot \} $$ is not monotonic: $F(\{t\}) = \{\bot,\mathrm{id}_{ \{ t \} }\} \not \subseteq F(\{t,f\}) = \{ \bot,\mathrm{id}_{\{t,f\}},\mathrm{not} \}$ if you represent $A \to B \subseteq A \times B$ as its graph (note that $\mathrm{id}_{\{t\}} \neq \mathrm{id}_{\{t,f\}}$; the former has a graph of one element while the latter has a graph with two elements), so we can't use the usual fixpoint theorems.
How can we show in the simplest possible terms (preferrably by counter-example) that a (the least) fixed-point of $F$ does in fact not exist? (Which is the result I would expect, but I'm prepared to be surprised.)
In hindsight this is probably somewhat obvious to an undergrad mathematician, but it is of vital importance to programming language semanticists, so here it goes:
Suppose that $μF$ is the fixed-point of $F$. Then there must be an injection $fun : (μF \to μF) \to μF$. On the other hand, $μF$ contains at least two elements (e.g., $\bot, (λ\_.\bot)$), so the cardinality of $μF \to μF = μF^{μF}$ is at least the cardinality of $2^{μF}$, the set of two-valued functions on $μF$. But $2^{μF}$ is one-to-one with the powerset $\mathcal{P}(μF)$, and Cantor has proven that the latter has strictly larger cardinality than $μF$ and then so has $μF^{μF}$. That is in contradiction to the existence of the injection $fun$.