New answers tagged set-theory
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 ...
5
votes
Accepted
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: $\...
2
votes
Accepted
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 $D \subseteq \omega$...
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 ...
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 ...
6
votes
Accepted
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\...
3
votes
Accepted
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 ...
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 ...
1
vote
Accepted
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
Accepted
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 ...
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(\...
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)>\...
5
votes
Accepted
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 ...
4
votes
Accepted
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 $\aleph_1$ ...
4
votes
Accepted
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 ...
14
votes
Accepted
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 $\...
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 $R=\{a,b\}$.
6
votes
Accepted
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 ...
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 ...
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{...
5
votes
Accepted
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 $M_2=M_1\cup\{\frac1n:n\in\mathbb N\}$ ...
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 ...
3
votes
Accepted
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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