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The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and as such carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form the paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;(*)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we can derive $\bot$. Therefore, $p \in P \rightarrow \bot$.

On the other hand formula $(*)$ holds for every $y$ --- substituting $p$ by $j(P)$ we have to show: $$p \in y \rightarrow j(P) \not= j(y)$$ or equivalently: $$p \in y \wedge j(P) = j(y) \rightarrow \bot$$ since $j$ is injective we have $j(P) = j(y) \rightarrow P = y$ and by composing it with the proof of $p \in P \rightarrow \bot$ we get the above formula. Thus, $p \in P$, which together with $p \in P \rightarrow \bot$ leads to the contradiction.

This means that there are can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ --- there is an obvious injection $A \rightarrow \Omega^A$ and composing injections gives injection. Thus, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{(-)}}$.

More generally, by a similar argument, one may show that if there exists an injection $\Omega \rightarrow X$ then there could be no initial algebra for $X^{X^{(-)}}$.

The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and as such carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form a paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;\(F\)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we have: $$p \in P \rightarrow p \not= j(P)$$ and by using (again) the assumption $p \in P$, we can derive $p \not= J(P)$, which by the definition of $p$ produces $\bot$. Therefore we constructed a method of turning a statement $p \in P$ into absurd, that is $p \in P \rightarrow \bot$.

On the other hand, we may show that the formula (F) holds for every $y$. By the definition of $p$, it is equivalent to: $$p \in y \rightarrow j(P) \not= j(y)$$ and by the definition of the implication, to: $$j(P) = j(y) \overset{\psi \circ \phi}\rightarrow (p \in y \rightarrow \bot)$$ Now, we may use our extra assumption saying that $j$ is injective: $$j(P) = j(y) \overset{\phi}\rightarrow P = y$$ and cut it with: $$P = y \overset{\psi}\rightarrow (p \in y \rightarrow \bot)$$ which is equivalent to: $$P = y \wedge p \in y \rightarrow \bot$$ and holds because $p \in P \rightarrow \bot$ as has been shown in the first part of the proof. Therefore, (F) holds as the composition of proofs $\psi$ with the fact $\phi$ saying that $j$ is injective. Finally, by comprehension, $p \in P$. So: $$(p \in P) \wedge (p \in P \rightarrow \bot)$$ thus: $$\bot$$

This means that there can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ (because we have an obvious injection $A \rightarrow \Omega^A$ and composition of injections is an injection). Therefore, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{\(-\)}}$.

More generally, by a similar argument, one may show that if there exists an injection $\Omega \rightarrow X$ then there could be no initial algebra for $X^{X^{(-)}}$.

The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form the paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;(*)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we can derive $\bot$. Therefore, $p \in P \rightarrow \bot$.

On the other hand formula $(*)$ holds for every $y$ --- substituting $p$ by $j(P)$ we have to show: $$p \in y \rightarrow j(P) \not= j(y)$$ or equivalently: $$p \in y \wedge j(P) = j(y) \rightarrow \bot$$ since $j$ is injective we have $j(P) = j(y) \rightarrow P = y$ and by composing it with the proof of $p \in P \rightarrow \bot$ we get the above formula. Thus, $p \in P$, which together with $p \in P \rightarrow \bot$ leads to the contradiction.

This means that there are can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ --- there is an obvious injection $A \rightarrow \Omega^A$ and composing injections gives injection. Thus, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{(-)}}$.

The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and as such carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form the paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;(*)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we can derive $\bot$. Therefore, $p \in P \rightarrow \bot$.

On the other hand formula $(*)$ holds for every $y$ --- substituting $p$ by $j(P)$ we have to show: $$p \in y \rightarrow j(P) \not= j(y)$$ or equivalently: $$p \in y \wedge j(P) = j(y) \rightarrow \bot$$ since $j$ is injective we have $j(P) = j(y) \rightarrow P = y$ and by composing it with the proof of $p \in P \rightarrow \bot$ we get the above formula. Thus, $p \in P$, which together with $p \in P \rightarrow \bot$ leads to the contradiction.

This means that there are can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ --- there is an obvious injection $A \rightarrow \Omega^A$ and composing injections gives injection. Thus, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{(-)}}$.

More generally, by a similar argument, one may show that if there exists an injection $\Omega \rightarrow X$ then there could be no initial algebra for $X^{X^{(-)}}$.

The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form the paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;(*)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we can derive contradiction $\bot$. Therefore, $p \in P \rightarrow \bot$.

On the other hand formula $(*)$ holds for every $y$ --- substituting $p$ by $j(P)$ we have to show: $$p \in y \rightarrow j(P) \not= j(y)$$ or equivalently: $$p \in y \wedge j(P) = j(y) \rightarrow \bot$$ since $j$ is injective we have $j(P) = j(y) \rightarrow P = y$ and by composing it with the proof of $p \in P \rightarrow \bot$ we get the above formula.

This means that there are can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ --- there is an obvious injection $A \rightarrow \Omega^A$ and composing injections gives injection. Thus, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{(-)}}$.

The above answers are great, but I would like to stress on one fundamental aspect here.

Contrary to some common beliefs, Cantor's diagonal argument is purely constructive and carries to the internal logic of elementary topos (notice, however, that it relies on impredicativity of the topos). For let us assume, there is an injection $j \colon \Omega^A \rightarrow A$. We may form the paradoxical subset of $\Omega^A$: $$P = \{x \in A \colon \forall_{y \in \Omega^A} x \in y \rightarrow x \not= j(y) \}$$ Let us consider: $$p = j(P)$$ If $p \in P$ then according to the definition of $P$: $$p \in y \rightarrow p \not= j(y) \;\;\;\;\;\;(*)$$ for all $y \in \Omega^A$, so particularly for $y = P$, we can derive $\bot$. Therefore, $p \in P \rightarrow \bot$.

On the other hand formula $(*)$ holds for every $y$ --- substituting $p$ by $j(P)$ we have to show: $$p \in y \rightarrow j(P) \not= j(y)$$ or equivalently: $$p \in y \wedge j(P) = j(y) \rightarrow \bot$$ since $j$ is injective we have $j(P) = j(y) \rightarrow P = y$ and by composing it with the proof of $p \in P \rightarrow \bot$ we get the above formula. Thus, $p \in P$, which together with $p \in P \rightarrow \bot$ leads to the contradiction.

This means that there are can be no injection $\Omega^A \rightarrow A$ for any $A$. This also means that there can be no injection $\Omega^{\Omega^A} \rightarrow A$ --- there is an obvious injection $A \rightarrow \Omega^A$ and composing injections gives injection. Thus, there are no isomorphisms $\Omega^{\Omega^A} \approx A$ and by Lambek's theorem, there are no initial (nor final) (co)algebras for $\Omega^{\Omega^{(-)}}$.