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The question is what do you mean by large cardinals and what do you mean by useful? For example, it is consistent that there exists a set $A$ such that the cofinite filter on $A$ is an ultrafilter. We can even show that such ultrafilter is closed under intersection of ordinal-indexed sequences (mainly because under this assumption every such sequence is ...


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The problematic part is that if $\kappa$ is singular, there might not be any definable first-order function witnessing this. So the only way to find it is essentially to just take the entire power set of $V_\kappa$, and hope to find a short cofinal sequence. If, however, you replace the schema of Replacement by the second-order statement "For every function ...


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No, it is not provable. Quite contrary, it is provably false. More precisely: A cardinal $\kappa$ is worldy iff $V_{\kappa} \models \operatorname{ZFC}$. We can prove that the least worldly cardinal (if any exist) has cofinality $\omega$ and hence is far from being inaccessible.


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I think that whenever I saw anyone using anything stronger than $\sf ZFC$, the axioms were defined explicitly. Except, perhaps, things like $I0,I1,I2,I3$ which are essentially the conventional names for these statements. Most of the time you write either something like "We abbreviate by $\sf IC$ the statement "There exists an inaccessible cardinal", and in ...


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HINT: Let $f:\kappa\to C$ be the canonical increasing enumeration, and let $F=\{\alpha\in\kappa:f(\alpha)=\alpha\}$. Show that $F$ is a club in $\kappa$. Let $K$ be the set of cardinals in $\kappa$. Show that $K$ is a club in $\kappa$. Show that $C'=C\cap F\cap K$.


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Some objects whose cardinal is at least $2^c,$ where $c$ is the cardinal of the reals : (1). The set of all real functions. (2). The set of all filters on an infinite set. (3). The dual space $l_{\infty}^*$ of the Banach space $l_{\infty}.$ (4). The maximal compactifications of $N$,of $Q$,and of $R$. (5). The free group on any set $S$ such that $|S|>c.$ ...


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A Hilbert space is a set with some structure. It's not hard to show in ZFC that Hilbert spaces exist, but if we wanted to we could consider some extremely weak theory $T\subset ZFC$ which can't prove e.g. that the reals form a set; then the statement "There is a Hilbert space" might not be provable in $T$, and would - if we wanted to use Hilbert spaces - ...


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There are two separate questions here: What are some "large" naturally occurring (that is, outside logic :P) mathematical objects? What are some applications of large cardinals which live outside logic? These are really separate questions, and the former has beed addressed elsewhere on this network (see e.g. ...


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Thanks for your kind words about my blog post. Let me try to answer your question. To describe the consistency strength of a theory or assertion, we should compare the consistency of that theory or assertion to that of other more familiar or landmark theories or assertions. For example, the consistency strength of ZFC plus the continuum hypothesis is ZFC ...


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Your language is countable, your structure is well-orderable. This means that you can prove the existence of Skolem functions without appealing to choice. So the usual proof should work pretty much out of the box.


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This post on the other math forum goes into more detail, as regards the metamathematical theory (it's subtle!). The mentioned rank initial segment $V_\kappa$ are all sets with rank $< \kappa$ (the inaccessible), where every set has a rank by the axiom of regularity in ZFC. See wikipedia with its link to the cumulative hierarchy. If you think of all sets ...


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In $V_\kappa$ it is very easy to understand what is $\kappa$. It's the class of all the ordinals. So the sentence is really "For every relation over $V_\kappa$ which is a function with domain being an ordinal, the range is bounded". For every relation over $V_\kappa$, since if $\kappa$ were singular, then a function witnessing that is not an element of ...



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