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Questions tagged [ordinals]

In the ZF set theory ordinals are transitive sets which are well-ordered by $\in$. They are canonical representatives for well-orderings under order-isomorphism. In addition to the intriguing ordinal arithmetics, ordinals give a sturdy backbone to models of ZF and operate as a direct extension of ...

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How do you define ordinal exponentiation without induction?

I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ...
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Proof of a theorem relating to ordinals [duplicate]

Please help me to prove this: If $X$ is any ordinal, then either $X$ is a limit ordinal or else there is a limit ordinal $Y$ and a natural number $n$ such that $X=Y+n$.
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30 views

Proving $\alpha+\beta=\sup\{\alpha+\beta_{\delta}:\delta<\gamma\}$

For ordinals $\alpha$, $\beta$, $\gamma$, if $\gamma$ is a limit ordinal and $\beta = \sup\{\beta_{\delta}:\delta<\gamma\}$, why does below expression hold, $$\alpha+\beta=\alpha + \sup\{\beta_{\...
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1answer
24 views

If $\aleph_\alpha=\alpha$, then $\alpha$ is a limit ordinal

If $\aleph_\alpha=\alpha$, then $\alpha$ is a limit ordinal. My attempt: Assume the contrary that $\alpha$ is not a limit ordinal. Then $\alpha$ is a successor ordinal and thus $\alpha=\beta+1$ for ...
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41 views

Boolean algebras/ Shelah/ Unclear step in the proof

Here on the page $10$, what does the displayed formula in the $6$th line $$\text {lg}(\eta_\ell)<\omega\Rightarrow \bigcup\{\text{Rang}(\nu_\ell(k):k<\text{lg}(\nu_\ell))\}\cap\bigcup_{k<\...
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1answer
60 views

There are arbitrary large singular cardinals $\aleph_\alpha$ such that $\aleph_\alpha=\alpha$

Definitions: Let $(\alpha_\xi \mid \xi < \lambda)$ be a transfinite sequence of ordinals of length $\lambda$. We say that the sequence is increasing if $\alpha_\nu < \alpha_\mu$ whenever $\nu&...
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1answer
33 views

If $(A,\preccurlyeq)$ is a linear ordering such that $|\{y\in A\mid y\preccurlyeq x\}| < \aleph_\gamma$ for all $x\in A$, then $|A|\le\aleph_\gamma$

If $(A,\preccurlyeq)$ is a linear ordering such that $|\{y\in A\mid y\preccurlyeq x\}| \le \aleph_\gamma$ for all $x\in A$, then $|A|\le\aleph_\gamma$. Does my attempt look fine or contain logical ...
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20 views

Well-Ordering Principle implies Zorn's Lemma

Well-Ordering Principle implies Zorn's Lemma Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My attempt: Let $(A,\...
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31 views

Axiom of Choice implies Well-Ordering Principle

Axiom of Choice implies Well-Ordering Principle. My textbook only presents the construction of function $F$ and does not provide details on how to define such well-ordering. I try to fill in the ...
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2answers
75 views

Is there a “solution” to the ordinal game?

Even though I have almost no background in logic, I find the idea of ordinal notation quite interesting. It seems that the idea is to come up with notation to define larger and larger numbers, until ...
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Let $α$ and $β$ be ordinals. Let $(η_{ζ})_{ζ\inβ}$ be a $β$-sequence of ordinals. Then $α\times(\sum_{ζ\inβ}\eta_{ζ})=\sum_{ζ\inβ}(α\times\eta_{ζ})$.

Let $\alpha$ and $\beta$ be ordinals. Let $\left(\eta_{\zeta}\right)_{\zeta\in\beta}$ be a sequence of ordinals indexed by $\beta$. Define the sum $\sum_{\zeta\in\beta}\eta_{\zeta}$ as follows. Let $Z=...
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26 views

Is the function sending each set to its corresponding Hartogs number injective?

Let $V$ be the class of all sets, $\text{Ord}$ be the class of all ordinals, $\text{InOrd}$ be the class of all initial ordinals, and $f:V \to \text{Ord}$ be the function that sends each set to its ...
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19 views

The Hartogs number of $A$ exists for any set $A$

The Hartogs number of $A$ exists for any set $A$. Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My attempt: Let $H=\{(W,\...
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50 views

For any set $A$, the Hartogs number of $A$ is an initial ordinal

The Hartogs number of $A$ is the least ordinal which is not equipotent to any subset of $A$. We denote the Hartogs number of $A$ by $h(A)$. Theorem: For any set $A$, $h(A)$ is an initial ordinal. ...
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1answer
25 views

Each well-orderable set $X$ is equipotent to a unique initial ordinal

Each well-orderable set $X$ is equipotent to a unique initial ordinal. Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My ...
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0answers
28 views

Let $n,k\in\omega$. Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$

Let $n,k\in\omega$. Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$. Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly ...
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24 views

Is it true that the class of initial ordinals is proper?

It is well-known that the class of ordinals is the proper class. Is it true that the class of initial ordinals is proper? Thank you for your help!
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1answer
70 views

Prove that $|\omega+\omega|=|\omega\cdot\omega|=|\omega^\omega|=|\omega|$

$|\omega+\omega|=|\omega\cdot\omega|=|\omega^\omega|=|\omega|$ Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help! My attempt: ...
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1answer
62 views

$\omega_2$ is not the countable union of countable sets [closed]

I'm not sure I quite understand the form of the proof in this post '$\omega_2$ is a not countable union of countable sets without AC' and similar ones. Is the idea to firstly show that there is an ...
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31 views

The countable union of countable ordinals is countable, using ZF [duplicate]

My understanding is that the more general case of this (the countable union of countable sets) requires choice to prove. However in the case of ordinals, it should be provable without choice since ...
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Making sense of a bijection between proper classes? [duplicate]

I am trying to solve the following problem in ZF: "Show that the collection $\{\kappa \ | \ \kappa=\aleph_\kappa\}$ is a proper class. My thinking is that in the construction any element of this ...
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1answer
61 views

Ordinal multiplication

I understand why $(\omega+1)\cdot2 = \omega\cdot2+1$ and why $(\omega+1)\cdot\omega = \omega^2$ What I am struggling with is something along the lines of: $(\omega\cdot3+4)\cdot3$ which I think is = ...
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1answer
51 views

Ordinal arithmetic with multiple brackets

So, I understand that multiplying ordinals has a distributive law on the left such that: $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$ What I am struggling with is if $\alpha$ is ...
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1answer
29 views

Is it true that for any cardinal $\kappa$ that in injects into $\aleph_{\omega}$, $(\kappa)^{+}$ injects into $\aleph_{\omega}$?

Here, $(\kappa)^+$ is the Hartog's Number, the least cardinal $\lambda$ so that there is no injection from $\lambda$ to $\kappa$. My feeling is that this is true, since for any ordinal $\alpha < \...
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1answer
35 views

Proof that an ordinal number does not contain itself [duplicate]

Hrbacek and Jech gives the following definition for ordinal numbers: However, the following proof seems to rely on the fact that an ordinal number does not contain itself (argument circled in red). ...
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2answers
50 views

Showing that $[0, \omega_1]$ is compact. [duplicate]

I want to show that $[0, \omega_1]$ is compact, where $\omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals. The tips I have seen to showing this ...
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1answer
31 views

A weak Goodstein sequence will eventually terminate

A weak Goodstein sequence starting at $m>0$ is a sequence $m_0,m_1,\cdots$ of natural numbers defined by $m_0=m$. To obtain $m_{k+1}$ form $m_k$ (as long as $m_k\neq 0$), write $m_k$ in base $k+2$, ...
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Why doesn't ZFC “know about” all of the ordinals?

My understanding is that there exist ordinals which ZFC cannot prove really are ordinals, i.e. that they are well-founded. I don't understand how this can be. Doesn't the iterative von Neumann ...
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1answer
64 views

Conway Notation for Large Countable Ordinals

I have not previously seen anything online that dives deeply into On: In Conway's notation On denotes the ordinal numbers (and No denotes the set of all surreal Numbers). Basically the elements of ...
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1answer
37 views

Every ordinal $\alpha>0$ can be expressed uniquely as $\alpha=\omega^{\beta_1}\cdot k_1+\omega^{\beta_2}\cdot k_2+\cdots+\omega^{\beta_n}\cdot k_n$

Every ordinal $\alpha>0$ can be expressed uniquely as $$\alpha=\omega^{\beta_1}\cdot k_1+\omega^{\beta_2}\cdot k_2+\cdots+\omega^{\beta_n}\cdot k_n$$ where $\beta_1>\beta_2>\cdots&...
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26 views

Let $\gamma,\alpha\neq 0$ be ordinals. Then there exists a unique ordinal $\beta$ and a unique $\rho<\alpha$ such that $\gamma=\alpha\cdot\beta+\rho$

Let $\gamma,\alpha\neq 0$ be ordinals. Then there exists a unique ordinal $\beta$ and a unique $\rho<\alpha$ such that $\gamma=\alpha\cdot\beta+\rho$. Does my attempt look fine or contain logical ...
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1answer
22 views

Let $\alpha,\gamma$ be ordinals such that $0<\alpha\le\gamma$. Then there is a greatest ordinal $\beta$ such that $\alpha\cdot\beta\le\gamma$

Let $\alpha,\gamma$ be ordinals such that $0<\alpha\le\gamma$. Then there is a greatest ordinal $\beta$ such that $\alpha\cdot\beta\le\gamma$. Does my attempt look fine or contain logical flaws/...
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1answer
54 views

Burali-Forti paradox for cardinals

It is well-known that in $\mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets). Is the same true for cardinals? (ordinals satisfying $(\forall \...
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1answer
45 views

Compositions of ordinal numbers

Let $\mathbf{Ord}$ denote the class of ordinals. Equipped with its natural (commutative) sum and product, this is a semiring. Say that a composition on $\mathbf{Ord}$ is a function $\circ: \mathbf{...
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1answer
64 views

Could there be an $\omega_1^{CK}$th hyperoperation?

If addition is the first hyperoperation, multiplication is the second, and the $(\alpha+1)$th hyperoperation is repeated occurrences of the $\alpha$th one. Is it possible for a limit ordinal (for ...
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1answer
120 views

Is $\omega + \omega^2 = \omega^2$ true?

Just a simple question. Is $\omega + \omega^2$ equal to $\omega^2$, I’ve just been thinking about it and if that’s false and it equals $\omega^2+\omega$ then we could define a set of countable ...
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29 views

Let $X$ be a set of ordinals. Then $\sup X\notin X\implies\sup X$ is a limit ordinal

Does my attempt look fine or contain logical gaps/mistakes? Any suggestion is appreciated. Thank you for your dedicated help! Let $X$ be a set of ordinals. Then $\sup X\notin X\implies\sup X$ is a ...
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24 views

Let $f$ be a mapping, $\beta$ be an ordinal, $X=\{\alpha\mid f(\alpha)\le \beta\}$, and $\gamma=\sup X$. Is $\gamma\in X$?

Let $f:\operatorname{Ord}\to\operatorname{Ord}$ be a mapping consisting of only addition, multiplication, and exponentiation operations, and $\beta$ be an ordinal. Let $X=\{\alpha\mid f(\alpha)\le \...
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67 views

How does countable collapse work?

As I’ve started delving deeper into the world of infinite ordinals, I’ve started seeing large, recursive, countable ordinals expressed in the form: $\varepsilon_{\kappa+1}$ where kappa is a large ...
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1answer
44 views

How do I proceed to prove $\tau<\alpha\cdot\beta\iff \tau=\alpha\cdot\eta+\zeta$ for a unique $\eta<\beta,\zeta<\alpha$?

Let $\alpha,\beta,\gamma,\tau$ be ordinals. Then $\tau<\alpha\cdot\beta\iff \tau=\alpha\cdot\eta+\zeta$ for a unique $\eta<\beta,\zeta<\alpha$ $\tau<\alpha^\beta\alpha^\gamma\...
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1answer
63 views

A question about the proof of $\sup\{\alpha^\beta\alpha^\delta\mid\delta<\gamma\}=\alpha^\beta\sup\{\alpha^\delta\mid\delta<\gamma\}$

I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here. Below is the verbatim proof by @...
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2answers
39 views

Let $\alpha,\gamma$ be ordinals and $A$ be a set of ordinals. Then $\gamma<\alpha+\sup_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$ [duplicate]

Let $\alpha,\gamma$ be ordinals and $A$ be a set of ordinals. Then $$\gamma<\alpha+\sup\limits_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$$ This problem arises when I prove ...
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2answers
53 views

Let $\alpha,\beta$ be ordinals. Then the lexicographic ordering of $\alpha\times\beta$ has order type $\beta\cdot\alpha$

Let $\alpha,\beta$ be ordinals. Then the lexicographic ordering of $\alpha\times\beta$ has order type $\beta\cdot\alpha$. This theorem comes from textbook Introduction to Set Theory by Hrbacek and ...
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1answer
46 views

A logical gap in the proof of $\sup\limits_{\beta\in A}(\alpha+\beta)=\alpha+\sup\limits_{\beta\in A}(\beta)$

I found this proof on a notes CONSTRUCTING THE INTEGERS: $\Bbb N$, ORDINAL NUMBERS, AND TRANSFINITE ARITHMETIC by Jeremy Booher. You can download freely it here. At page 6, he presents Lemma 28 and ...
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22 views

Let $\beta=\sup\limits_{\nu<\gamma}\beta_\nu$. How do the authors prove $\alpha+\beta=\sup\limits_{\nu<\gamma}(\alpha+\beta_\nu)$?

Here is an excerpt from my textbook Introduction to Set Theory by Hrbacek and Jech. Although I'm able to prove these three equalities, my proof is up to a page long containing even a lemma (If you are ...
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2answers
55 views

Let $α$ be an ordinal and $A$ be a set of ordinals. Then $\sup\limits_{β∈A} (α+β) = α+\sup\limits_{β∈A}(β)$

My idea is to prove that $\alpha+ \sup\limits_{β∈A}(β)$ is the supremum of $\{α+β\mid β∈A\}$. While I'm able to prove that $\alpha+ \sup\limits_{β∈A}(β)$ is a upper bound of $\{α+β\mid β∈A\}$, I ...
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1answer
58 views

Can the function $S(\alpha)$ have a fixed point ordinal?

I feel like there is no fixed point of the successor function, as for any ordinal $\alpha$, the ordinal $\alpha \cup \{\alpha\}$ must be greater. Though maybe I’m wrong and some ordinals with ...
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2answers
36 views

How to prove $\sup\limits_{\delta<\gamma}(\alpha+(\beta+\delta))=\sup\limits_{\epsilon<\beta+\gamma}(\alpha+\epsilon)$

I'm trying to prove this theorem but stuck at proving $\color{blue}{\epsilon<\beta+\delta'\text{ for some }\delta'<\gamma\implies\epsilon=\beta+\delta\text{ for some }\delta<\gamma}$. Any ...
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1answer
54 views

The reasoning behind $\sup\{\alpha\beta+\alpha\zeta\mid\zeta<\gamma\}=\alpha\beta+\sup\{\alpha\zeta\mid\zeta<\gamma\}$

Let $\alpha,\beta,\gamma$ be ordinals. Then $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$. I'm been looking for some proofs of this theorem on MSE and found that the gist is the equality $\sup\{\...
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22 views

If $\alpha\le\beta$, then there is a unique ordinal $\gamma$ such that $\alpha+\gamma=\beta$

Does my attempt look fine or contain logical gaps and flaws? I'm grateful for your help! If $\alpha\le\beta$, then there is a unique ordinal $\gamma$ such that $\alpha+\gamma=\beta$. My attempt: ...