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In abstract category theory morphisms are not required to be functions: we are given an arbitrary class, regarding its elements as arrows, equipped with domain and codomain information and an associative operation. If we want, we can specify so that for a funtion $f:A\to B$, let ${\rm dom\,}f:=B$ and ${\rm cod\,}f:=A$ and the operation is defined as ...


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You can't construct such example. The fact that it is only consistent and not provable is exactly the issue here. Not only you can't construct one, you can't prove that one exists when the axiom of choice fails. These remarks, at the introductory level, are meant to guide the reader and suggest that a choice-free proof cannot be found. This is used to both ...


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For the $\omega_1$ case, the reason is that at limit stages if we collect all the branches, we'll have uncountably many points, so we have to allow only a few branches to go through limit steps. As for the second question, $\sf CH$ gives us exactly that $\aleph_1^{<\aleph_1}=\aleph_1^{\aleph_0}=\aleph_1$, which is why there is an $\aleph_2$-Aronszajn ...


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Let $A$ be the biggest of the two $$ |A|\le|A\times B|\le|A^2| $$ The last piece you miss is $|A|=|A^2|$, that is equivalent to Choice Axiom, and can be proved, for example, through Zorn's Lemma


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This requires the axiom of choice in all accounts. First because the axiom of choice is equivalent to the fact that every two cardinals can be compared (otherwise $\max$ has no meaning); and secondly because we use the fact that $|A|^2=|A|$ for infinite sets, which is also equivalent to the axiom of choice. Now, using these two facts we have that if ...


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Let me begin with two examples. The first example is the famous Hilbert and Brouwer controversy. A part of this philosophical contest was on Cantor's theory of infinite numbers. Of course Brouwer as an intuitionist is against using infinite numbers in maths. On the other side we have Hilbert's famous sentence that "No one can expel us from Cantor's ...


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No, $\beta_0^\alpha=\beta$, so $\beta_1^\alpha$ is the ordinal $\min(C_\beta\setminus\alpha)$: it’s the smallest member of $C_\beta$ that is greater than or equal to $\alpha$. It exists because $C_\beta$ is cofinal in $\beta$, and it’s necessarily less than $\beta_0^\alpha=\beta$, since $C_\beta\subseteq\beta$. If this is greater than $\alpha$, repeat the ...


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There is exactly one morphism from $X$ to $\emptyset$, namely the unique function from $\emptyset$ to $X$.


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I think the problem should say: Construct $\mathcal{B} \subseteq \mathcal{A}$, such that ... The hint is saying that letting $\mathcal{B} = \{b_{\alpha} : \alpha < \omega_1\}$, $\mathcal{A} - \mathcal{B} = \{a_{\alpha} : \alpha < \omega_1\}$ should work where $\mathcal{A}$ is a.d. family and for all $\alpha \neq \beta$, $a_{\alpha} \cap b_{\alpha} = ...


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Perhaps the father of modern infinite deserves to be quoted here: My theory stands as firm as a rock; every arrow directed against it will quickly return to the archer. How do I know this? Because I have studied it from all sides for many years; because I have examined all objections which have ever been made against the infinite numbers; and ...



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