# Tag Info

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I am assuming you know that $|\Bbb R|=2^{\aleph_0}$, which can be proven by looking at binary expansions of numbers in $[0,1]$ (discounting countably many numbers with non-unique expansions). The cardinal $\aleph_1$ is by definition the smallest cardinal larger than $\aleph_0$, meaning that there is no set $A$ such that $\aleph_0<|A|<\aleph_1$. Thus ...

5

A nonprincipal ultrafilter on $\mathbb N$ does not consist of the set of all cofinite sets, but it includes the set of all cofinite sets. In order to be an ultrafilter and not just any filter it must be a filter that is not a subset of any other filter. That means, for example, that if neither the set of even numbers nor the set of odd number belongs to ...

5

No. For two main reasons: Even in the absence of choice $\aleph_\omega$ cannot be a power set of anything. The proof is the same usual proof that relies on Koenig's theorem, which may require choice, but it does not require choice if the power set is well-orderable. So in any case it is provable you will have uncountable cardinals which are not power ...

3

I claim that more than ZF is needed for this result. Specifically, I claim that, in the basic Fraenkel model, the set $A$ of atoms with the discrete topology is a counterexample. (I'm using a permutation model here, but the result transfers to ZF by the Jech-Sochor theorem.) Of course, $A$ isn't compact because the cover by singletons has no finite ...

3

Let us assume to avoid trivialities that $U$ is required to contain no sets of cardinality smaller than $|I|$. Even with this assumption, no matter how large $I$ is, your desired conclusion will not hold for arbitrary ultrafilters. Indeed, let $I$ be any infinite set and partition $I$ into countably many sets $I_n$ with the same cardinality as $I$. Define ...

3

$A\cap B=\{x\in A\mid x\in B\}$. So intersection can be deduced from comprehension axiom scheme.

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$f$ does not occur in the formal version. In fact, it is the function defined by the propositional function $P(y,z)$. That is, $P$ gives us an instance of the Axiom Schema of Replacement only if the first condition about $P$ is met (is a theorem). In that case, our $f$ is just the function that maps $x$ to the unique $y$ for which $P(x,y)$ holds. In other ...

3

Assuming the axiom of choice, every two cardinals are comparable. In particular either $\aleph_1\geq 2^{\aleph_0}$ or $\aleph_1\leq 2^{\aleph_0}$. Since $\aleph_1$ is the least uncountable cardinal, and $2^{\aleph_0}$ is uncountable by the diagonal argument, it follows that it is necessarily the case that $\aleph_1\leq 2^{\aleph_0}$. So the continuum ...

3

If by "what does it mean" you mean "what does it mean" I have to ask what do you mean, what does it mean? Saying that ZF has infinitely many axioms means exactly what it says. There are infinitely many primes. There are only finitely many natural numbers less than $3$. There are infinitely many axioms of ZF. Probably you actually understand that, and you're ...

2

No. The axiom already implies that there's only one such $x$. Given $y$, let $x$ and $x'$ be elements with ${\forall}z(P(y,z){\iff}(x=z))$ and ${\forall}z(P(y,z){\iff}(x'=z))$. Using $z=x$ in the first statement yields $P(y,x)\iff(x=x)$, and thus $P(y,x)$; then using $z=x$ in the second statement yields $P(y,x)\iff(x'=x)$, and thus $x'=x$.

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$\aleph_1$ is the least uncountable cardinal, and $|\mathbb{R}| = 2^{\aleph_0}$. To say that there is no uncountable set with cardinality less than that of $\mathbb{R}$ is precisely to say that $|\mathbb{R}|$ is the least uncountable cardinal; that is, $2^{\aleph_0} = \aleph_1$.

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The idea is that you are defining $f(y)$ to be the unique $x$ such that $P(y,x)$ is true. If it said simply ${\forall}y({\exists}x:P(y,x))$ then you would be right that it needs to say $\exists!$ instead of just $\exists$. But notice that it says not just that there is an $x$ such that $P(y,x)$ but that there is an $x$ such that $P(y,z)\Leftrightarrow z=x$ ...

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The existence of intersection follows from the axiom schema of separation which informally tells you that if you start with a set and specify appropriately a way to select certain elements from this set by describing a formula that they satisfy then indeed there exists a subset containing precisely those elements. The construction of intersection fits this ...

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HINT: For the first one let $h:\omega_1\times\omega_1\to\omega_1$ be a bijection. Let $C=\{\alpha\in\omega_1:h[\alpha\times\alpha]=\alpha\}$, and show that $C$ is a club set in $\omega_1$. Then consider the sets $h^{-1}[A_\alpha]$ for $\alpha\in C$. For the second, let $h$ and $C$ be as above. For $\alpha\in C$ let $$f_\alpha=\begin{cases} \{\langle\xi,\... 1 Let me elaborate on Asaf's claim that \aleph_\omega cannot be a power set. Suppose |P(X)|=\aleph_\omega for some set X. Since |X|<|P(X)|, |X|<\aleph_\omega, and so |X|=\aleph_n for some n<\omega. It would then follow that$$\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{\aleph_n}=(2^{\aleph_n})^{\aleph_n}=2^{\aleph_n^2}=2^{\aleph_n}=\...

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Let $E=\{2n:n\in\Bbb N\}$, and let $\mathscr{B}=\{B_x:x\in\Bbb R\}$ be an almost disjoint family of infinite subsets of $E$; you’ll find constructions in the answer to this question and the one to which it’s linked as a duplicate. For $x\in\Bbb R$ let $A_x=B_x\cup(\Bbb N\setminus E)$; then $\mathscr{A}=\{A_x:x\in\Bbb R\}$ has the desired properties.

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What you’re missing, I think, is that as a set the tree $T$ is simply $\omega_1$: its elements are the countable ordinals. Fix $\alpha_0<\omega_1$. Given $\alpha_n<\omega_1$ for some $n\in\omega$, there is an $\alpha_{n+1}$ such that $\alpha_n<\alpha_{n+1}<\omega_1$, $\alpha_n\subseteq T\upharpoonright\alpha_{n+1}$, and \$T\upharpoonright \...

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The idea is that a coloring of pairs is equivalent to a graph (by considering the coloring as the characteristics function of the graph relation); and homogeneous sets are either a clique or an anti-clique. If your ultrafilter is Ramsey, then given any coloring you look at the induced graph, and you look at its connected components: either one of them is in ...

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