# Tag Info

5

The answer is no. It is known that ZF+'$\aleph_1$ and $2^{\aleph_0}$ are incomparable' is consistent if ZF does. Assume that $\aleph_1$ and $2^{\aleph_0}$ are incomparable (over ZF),then $\aleph_1$, $2^{\aleph_0}<\aleph_1+2^{\aleph_0}$ and $\aleph_1+2^{\aleph_0}\neq\aleph_1\cdot 2^{\aleph_0}$. 1 is trivial. To prove 2, we can use following theorem: ...

4

This is a generalization of tetori's nice answer, which in fact shows that this assumption implies the axiom of choice. Lemma I. If $a$ and $b$ are infinite cardinals equipollent with their own squares, then $(a+b)^2=a\cdot b$. Proof. We calculate: $$(a+b)^2=a^2+b^2+2\cdot a\cdot b=a^2+b^2+a\cdot b=a+b+a\cdot b=a\cdot b$$ The second equality comes from ...

4

The proof is simple enough that you don't need a book. You just prove it by induction. The base case (one set) is trivial, and the inductive case is not much harder, using the fact that $$\prod_{i = 1}^{n+1} X_i \cong \left (\prod_{i=1}^n X_i\right ) \times X_{n+1}.$$

3

This does not answer the question (which I misread, being in a mental fog), but I am undeleting it at the OP's request. The following summary is copied from p. 174 of Wacław Sierpiński, Cardinal and Ordinal Numbers, second edition revised, Warszawa, 1965: If, for a natural number $k$ and cardinal numbers $m$ and $n$, we have $km=kn$, then $m=n$. This ...

3

Let $\langle X,\le\rangle$ be a non-empty linear order. If $X$ has a maximum element $x_0$, then $\{x_0\}$ is a well-ordered cofinal subset, so assume that $X$ has no largest element. Let $$\mathscr{W}=\{W\subseteq X:\langle W,\le\rangle\text{ is a well-order}\}\;,$$ and define a relation $\preceq$ on $\mathscr{W}$ by setting $W_0\preceq W_1$ if and only ...

3

You want the Hartogs number of the set $X$: it’s the least well-ordered cardinal that can’t be injected into $X$, and its existence can be proved in $\mathsf{ZF}$. Briefly, if $X$ is a set, so is $$\mathscr{W}=\{\langle S,\preceq\rangle\in\wp(X)\times\wp(X\times X):S\subseteq X\text{ and }\preceq\text{ well-orders }X\}\;.$$ For each $\langle ... 3 No. You're absolutely wrong when you say that the second argument uses the fact that$A$"has a cardinality" The axiom of choice makes sure that every cardinality is an aleph number, but cardinality and its basic arithmetics have absolutely nothing to do with the axiom of choice. Every set has a cardinality, because cardinality of a set$A$is just the ... 3 In this answer I showed how to construct an independent family$\mathscr{I}$of subsets of$\Bbb N$of cardinality$2^\omega=\mathfrak{c}$. That is,$\mathscr{I}$has the property that $$\bigcap_{A\in\mathscr{J}}A\cap\bigcap_{A\in\mathscr{K}}(X\setminus A)\ne\varnothing$$ whenever$\mathscr{J}$and$\mathscr{K}$are disjoint finite subsets of ... 3 It isn't generally true that$\text{WO}(A)/\sim$has the same cardinality as$P(A)$; this would be true under the GCH and indeed the assertion that it is true for all infinite$A$is equivalent to the GCH in ZFC. Meanwhile, there is no injection from$\text{WO}(A)/\sim$to$A$, because if there were, you could use the image of that injection to make an ... 3 HINT: For$[U],[V]\in\operatorname{WO}(A)/\!\!\sim$define$[U]\preceq[V]$iff$\langle U,\le_U\rangle$is order-isomorphic to an initial segment (not necessarily proper) of$\langle V,\le_V\rangle$. Prove that$\preceq$is a well-defined well-ordering of$\operatorname{WO}(A)/\!\!\sim$. Now suppose that$h:\operatorname{WO}(A)/\!\!\sim\,\to A$is an ... 2 First note that for initial ordinals (an initial ordinal is an ordinal which is not equipotent to any smaller ordinal, e.g.$\omega$or$\omega_1$) being smaller as ordinals is the same as being smaller in cardinality. You can't quite prove this statement if you don't add the requirement that$\chi(A)$is an initial ordinal, which is the same as saying that ... 2 Yes, you've made a mistake. If you map everything to$0$, and there were$x,y$such that$x\lesssim y$but$y\not\lesssim x$, then it can't be that$\pi(x)=\pi(y)=0$. The idea is that if you have a preorder, then there is a natural quotient which gives a partial order. And a prewellordering is such where the quotient is a well-ordering. And the key point ... 2 If a generic set$G\subset P$were definable, then so would be its complement$P-G$. But this is a dense set, since every condition has incompatible extensions, and not both of them can be in$G$. So we'd have a definable dense set having no condition in$G$, which contradicts genericity. 2 By transfinite recursion, we can define a sequence$(a_\alpha)_{\alpha\in\mathrm{Ord}}$of elements of$A$(see ordinals) such that$a_\alpha$satisfies $$\forall\beta,\ \beta<\alpha\implies a_\beta<a_\alpha$$ Each step is possible if and only if$(a_\beta)_{\beta<\alpha}$isn't cofinal. But this construction must stop somewhere for there isn't an ... 2 Your definition is correct and here is the proof you are looking for: Let$\alpha$be an ordinal and let$\beta \in \alpha$. We show that$\beta$has the two properties you mentioned: Well-ordering: Let$X$be a non-empty subset of$\beta$. From the transitivity of$\alpha$, it follows that$X \subseteq \alpha$and since$\alpha$is well-ordered,$X\$ must ...

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