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1 answer
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Is $\operatorname{rank}(A)\subseteq\operatorname{TC}(A)$?

This question came to me when I was thinking about rank and transitive closures. Let $A$ be a set of rank $\alpha$ and let $\operatorname{TC}(A)$ denote the transitive closure of $A$. Is it true then ...
Anon's user avatar
  • 448
2 votes
1 answer
85 views

Can $\sf{ST}$ construct an infinite class wellordered ordered by $\in$?

Assume the axioms of Extensionality, Empty Set, and Adjunction (meaning that $S\cup\{x\}$ forms a set for any $S,x$). Notice that we do not have Specification as an axiom, which makes this theory very ...
Jade Vanadium's user avatar
1 vote
1 answer
83 views

Can cardinality $\kappa$ exist where $\forall n\in\mathbb{N} \beth_n<\kappa$,$\kappa<|\bigcup_{n\in\mathbb{N}}\mathbb{S}_n|$,$|\mathbb{S}_n|=\beth_n$

The Wikipedia article on Beth numbers defines $\beth_\alpha$ such that $\beth_{\alpha} =\begin{cases} |\mathbb{N}| & \text{if } \alpha=0 \\ 2^{\beth_{\alpha-1}} & \text{if } \alpha \text{ is a ...
SarcasticSully's user avatar
0 votes
0 answers
65 views

Can the class of ordinals be extended even further? [duplicate]

Is it possible for anything to come after all ordinals? I don't see why not. For example, one can take a non-ordinal set $S$, and then add in all the ordered pairs $(\alpha, S)$ to $ON$, where $\alpha$...
user107952's user avatar
  • 21.5k
2 votes
0 answers
73 views

Set of ordinals isomorphic to subsets of total orders

Background. Given a poset $(S,<)$ we'll indicate with $\tau(S,<)$ the set of all the ordinals which are isomorphic to a well ordered subset of $(S,<)$. We're in $\mathsf{ZFC}$. Questions. ...
lelouch_l8r4's user avatar
3 votes
1 answer
73 views

Constructible subsets of an ordinal - an alternate definition?

Work over ZFC. Recall the standard definition of $\mathrm{Def}(X)$, the set of constructible subsets of the set $X$. This can be considered as taking all those subsets of $X$ that satisfy some first-...
theHigherGeometer's user avatar
1 vote
0 answers
46 views

Ordinals multiplication: Does $a^2 b^2=b^2a^2$ imply $ab=ba$? [duplicate]

I found this question in one of set theory past exam: If $\alpha$ and $\beta$ are two ordinals such that $\alpha^2\beta^2=\beta^2\alpha^2$, does it necessarily imply $\alpha\beta=\beta\alpha$? Clearly ...
Ankiiatsy's user avatar
1 vote
1 answer
93 views

Intuition of Ordinal Number

I am learning the concept of ordinal numbers. In the book of Set Theory and Metric Spaces by I. Kaplansky (Sec. 3.2, pg. 55), the author states We attach to every well-ordered set an ordinal number; ...
Mingzhou Liu's user avatar
2 votes
0 answers
28 views

References that give the cofinality of ordinal addition, multiplication and exponentiation

$\newcommand{\cf}{\operatorname{cf}}$ Let $\alpha$, $\beta$ be ordinals. I believe that we have \begin{align*} \cf(\alpha+\beta)=\cf(\beta),\quad\beta\neq 0;\quad\cf(\alpha\cdot\beta)=\begin{cases}\cf(...
Jianing Song's user avatar
  • 1,923
2 votes
1 answer
104 views

Define the sum of transfinite ordinal sequences according to finite ordinal sum.

I know it is possibile to define finite sum of ordinals: if $(\alpha_i)_{i\in n}$ is a sequence of lenght $n$, with $n$ in $\omega$, then the symbolism $\sum_{i\in n}\alpha_i$ has a (formal) meaning; ...
Antonio Maria Di Mauro's user avatar
-2 votes
1 answer
57 views

Prove that the order type of $\alpha\cdot\beta$ is the antilexicographic order in $\alpha\times\beta$. [closed]

This question is related to this one, but not a duplicate, since I am struggling with injectivity and monotonicity, rather than proving that $\{\alpha\cdot\eta + \xi:\eta<\beta\textrm{ and }\xi<\...
Antonio Maria Di Mauro's user avatar
4 votes
1 answer
182 views

Colourful class function

Background. We're in $\mathsf{ZFC}$, and I can use the principle of $\epsilon$-induction, but not (directly) the $\epsilon$-recursion. Problem. Let $F : V \to V$, where $V$ is the class of all the ...
lelouch_l8r4's user avatar
0 votes
0 answers
75 views

How Should I Do the Inductive Case of this Proof on Ordinals?

Question Prove that for all ordinals $\alpha$, $V_\alpha=\{x:\rho (x)<\alpha\}$. Note: The function $\rho$ is a rank function. Attempt I did the proof by induction on ordinals. I started with the ...
Mr Prof's user avatar
  • 451
2 votes
0 answers
137 views

Strictly decreasing function from an ordinal to $\mathbb{R}$

Context: We are working in $\mathsf{ZFC}$. Problem: Given a poset $(P,<)$, let $\text{Dec}(P)$ be the set of all strictly decreasing function $f : \alpha \to P$, where $\alpha$ is an ordinal number....
lelouch_l8r4's user avatar
4 votes
1 answer
126 views

Is there a smarter way to calculate $F(\omega^{15} + 7)$? (Ordinal arithmetic)

Given the function $F : \text{Ord} \to \text{Ord}$ definite by transfinite recursion as it follows: $$F(0) = 0 \qquad F(\alpha + 1) = F(\alpha) + \alpha \cdot 2 + 1 \qquad F(\lambda) = \sup_{\gamma &...
lelouch_l8r4's user avatar

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