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2 votes

Ordering of large cardinals by cardinality

To elaborate on Joel David Hamkins's answer: When the the size order of large cardinal properties differs from the strength order and it is not an example of the identity crisis phenomenon, it is ...
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5 votes
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Existence of a particular function that maps an arbitrary set of ordinals to a single ordinal

The existence of a function $f$ as specified in the question cannot be proved in ZFC. This follows from the following theorem and the well-known independence of $\mathrm{V = OD}$ (equivalently: $\...
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2 votes
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Minimal cardinality of non-bipartite sub-family of $[\omega]^\omega$

The cardinal $\mathfrak{nb}$ is equal to the reaping number $\mathfrak{r}$. An unsplit family is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega$...
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4 votes

Can Category theory be founded in set theory using worldly cardinals instead of inaccessibles?

The title asks: Can Category theory be founded in set theory using worldly cardinals instead of inaccessibles? The first line in the main body of the question is: What is exactly demanded for a set ...
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5 votes

Can Category theory be founded in set theory using worldly cardinals instead of inaccessibles?

This ultimately depends on what 'kind' of category theory you want to do; it's one of those studies that can really get as 'large' as you want it to, and this is part of what enables us to 'study ...
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6 votes
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Weak form of $\text{CH}$ in $L(\mathbb{R})$

No, this can fail. Force $\mathrm{MA}_{\omega_1}$ over $L$ by ccc forcing and let $M$ be the $L(\mathbb R)$ of the extension $L[G]$. Note that $M$ has the same cardinals as $L$. We have $X=\mathbb R\...
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3 votes
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A continuous map relating co-constructible reals

By "formula" I will mean "formula of set theory with ordinal parameters." Note that "ordinal parameters" is equivalent to "constructible parameters" for our ...
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3 votes

Can countable ordinals start gaps of every order in the constructible universe?

Your other questions have been answered, so I provide here only an answer for your last edit, that is, whether an ordinal can be in a gap of every order. The answer is positive by an argument almost ...
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  • 361
1 vote
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Consistency strength of an attempt at higher order set theory

I believe $ZFC$ plus the existence of a countable collection of strictly inaccessible cardinals is also a lower bound on the consistency strength of this theory, and am posting this as a CW answer to ...
2 votes
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Optimal partitions amongst a given set of partitions

Modifying your note at the end slightly, try the following. Let $X=\{0,1,2,\ldots\}$ and let $P=\{\{0,1\},\{2,3\},\{4,5\},\ldots\}$. Let $\mathfrak{P}$ be all partitions of $X$ which refine $P$ and ...
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0 votes

Diagonalization over a normal function and its derivatives on transfinite ordinals

At least you have that $\Phi(\alpha,0)$ is indeed normal. It is increasing, because $\Phi(\alpha+1,0)$ is always greater than $\Phi(\alpha,0)$; and it is also continuous since, by definition, $\Phi(\...
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2 votes

Diagonalization over a normal function and its derivatives on transfinite ordinals

No, not all $G_\beta$ are normal. For example let $\Phi(0,\beta)$ be any normal function whose least fixed point is greater than $\omega$ and consider $G_\omega(\alpha)$. Since $\Phi(\beta+1,0)>\...
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  • 395
5 votes
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Given $\pi$ permutation on $\{1,\dotsc,n\}$, what is the sign of a permutation of $\{2,\dotsc,\hat\jmath,\dotsc,n\}$?

$\DeclareMathOperator\sgn{sgn}$This is almost the same as your previous question, just with the order of the operations switched—whether you think of $\pi$ as ordering or disordering is just a matter ...
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4 votes
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Can a stage of the cumulative hierarchy violate the partition principle?

If you are asking whether or not a $V_\alpha$ could violate the partition principle, the answer is easily yes. As we all know, it is always the case that $\Bbb R$ can be partitioned into $\aleph_1$ ...
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4 votes
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Sign of the permutation which brings a subsequence back to its original form

Imagine a bubble sort where you bring each element to its original position. $x_1$ would have taken $\pi^{-1}(1) - 1$ transpositions to bring it back to its original position. Let's perform those ...
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  • 7,933
14 votes
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Number of Laurent monomials of n variables with degree at most d

One such formula is $$\sum_{p=0}^n \binom{n}{p} \binom{d}{p} \binom{d+n-p}{n-p}.$$ To derive this, let $P \subseteq [n]$ be the set of variables with positive exponents and let $p = |P|$. There are $\...
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6 votes

MAD family with the choosability property

Take any MAD family on $\omega\setminus\{a,b\}$ whose intersection is $\varnothing$. Then add $a$ to some of its elements and $b$ to all other elements. Then you can choose $R=\{a,b\}$.
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6 votes
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MAD family with the choosability property

This answer only deals with the case that $R$ is infinite. I thought that I would be able to modify it to the finite case - thanks to Ilya Bogdanov for spotting the mistake in my argument. (His answer ...
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2 votes

Why can we assume a ctm of ZFC exists in forcing

This is to slightly elaborate on point 2 of Noah Schweiber's answer, since in my opinion this approach is often presented in a somewhat confusing manner which ommits some key subtelties. Forcing ...
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  • 951
17 votes

Why can we assume a ctm of ZFC exists in forcing

Expositionally, forcing is (usually) easier to understand with a c.t.m. This does indeed lead to somewhat different results, such as $(*)\quad$ If there is a countable transitive model of $\mathsf{...
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5 votes
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Product topology from two premetric spaces induced by sum of premetrics?

The answer to this question is negative. Consider the subspace $M_1=\{0\}\cup\{\frac 1n+\tfrac{i}{nm}:n,m\in\mathbb N\}$ of the complex plane and the space $M_2=M_1\cup\{\frac1n:n\in\mathbb N\}$ ...
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3 votes

Set theory without the empty set

I’ve explored a “set theory whose axioms do not prove the existence of an empty set,” the Incomprehensive Set Theory in my “Naive View of the Russell Paradox” (https://arxiv.org/abs/2103.00090), but ...
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3 votes
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Is every first-countable Lindelof space of cardinality $<\mathfrak c$ a $Q$-space under MA?

I think that the answer to both of your questions is negative. Let $X$ be a subspace of the real line with its usual topology of size $\omega_1$. Then, clearly, $X$ is first countable, Lindelöf and ...
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