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8

The cardinality of real numbers is indeed $2^{\aleph_0}$. However $\aleph_1$ is not defined as $2^{\aleph_0}$ (the cardinality of the continuum, often denoted $\mathfrak c$), but it is defined as the smallest cardinality strictly greater than $\aleph_0$. Whether $2^{\aleph_0}=\aleph_1$ (termed the continuum hypothesis) can neither been proven nor disproven ...


5

This is just not true. Suppose that $\sf ZFC+\operatorname{Con}(ZFC)$ is just inconsistent, and the additional assumption is the one causing the inconsistency. What does that mean? It means that in every model of $\sf ZFC$ it holds that $\lnot\operatorname{Con}\sf (ZFC)$, and therefore it is provable from $\sf ZFC$ that $\lnot\operatorname{Con}\sf (ZFC)$. ...


3

The axiom of choice was something people had used without always noting that there is an assumption to be made in order to justify making infinitely many choices at once. On the other hand, Cantor felt that there shouldn't be intermediate cardinals between the naturals and the reals, so he hypothesized that this is the case and spent a considerable amount ...


2

You should specify the order on your poset. But with the usual ordering where $p\leq q$ if $p\supseteq q$, yes, amoeba forcing is Knaster. To see this, fix a countable base for the topology and let $X$ be an uncountable subset of $\mathbb{P}$. By shrinking $X$ a bit we may also assume that there is an $n$ such that $\mu(p)\leq\varepsilon-\frac{1}{n}$ for ...


2

The simplest way, I think, is to use the following approach. We have the language, $\cal L$ which is an ordered tuple of pairwise disjoint sets, and a function $\sigma$. The sets are constants, relations, functions and variable symbols. And you have an arity function $\sigma$ which takes a relation and function symbol and returns a natural numbers for the ...


2

Let me first clarify the issue with an inaccessible, since it seems to me that my words were somewhat misunderstood. I only mentioned inaccessible cardinals, since it is a plausible axiom, and if you believe it to be consistent, then of course there will be a (standard) model of $\sf ZFC$. And in that case we cannot prove $\sf\lnot\operatorname{Con}(ZFC)$. ...


2

To summarize, and to answer the question, it seems that we have an unknown problem. The reason being: the OP has established "If $ZFC\nvdash$ $\neg$Con$(ZFC)$, then $ZFC\nvdash$ Con$(ZFC)$ $\implies SM$." in the statement of his question. Also, Asaf has established (the almost trivial) "If $ZFC\vdash$ $\neg$Con$(ZFC)$, then $ZFC\vdash$ Con$(ZFC)$ ...


2

This is not quite possible, but we can slightly fix this. First let me point out some defining characteristic of cardinals. These are well-orders that every proper initial segment has a strictly smaller cardinality. On the other hand, $\omega_2+\omega_1$ has an initial segment equipollent to the entire order. So now we run into a problem. If there is an ...


2

This argument is due to Sierpinski. Towards a contradiction, suppose $<_1$ is a Borel well ordering of reals. Let $r$ be the $<_1$-least real such that $\{x: x <_1 r\}$ is not null. The argument is essentially the same if no such $r$ exists. It follows that the set $W = \{(x, y): x <_1 r \wedge y <_1 r\} = A^2$ where $A = \{a: a <_1 r\}$ is ...


2

Actually, you don't need the Axiom of Foundation for that. Lemma. Given a set $X$, we can find a set $Y$ such that $|X|=|Y|$ and $X\cap Y=\emptyset.$ Proof. Let $$T=\{(S,x):S\subseteq X,\ x\in X,\ (S,x)\in X,\ (S,x)\notin S\}\subseteq X$$ and let $$Y=\{(T,x):x\in X\}.$$ Clearly $|X|=|Y|.$ Assume for a contradiction that $X\cap Y\ne\emptyset,$ i.e., there ...


1

Something is not clear to me about your question. Global choice implies the axiom of choice for sets. Of course it does. Every set is a subclass of $V$, and if $V$ can be well-ordered, then every set can be well-ordered. Requiring that there is an injection from the class of ordinals into $V$ is provable in $\sf ZF$. This function is the identity function. ...


1

This statement allready follows from the addition theorem, which states that $A \times \{0,1\}$ is equipotent with $A$ for every infinite $A$ (as on $A \times \{0,1\}$, $(a,x) \mapsto (a, 1+x)$ is fixed-point-free). The addition theorem is known to be strictly weaker than the axiom of choice (a result by Sageev).



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