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Results tagged with ordinal-numbers
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user 1946
An ordinal is the order type of a well-ordered set. The first few ordinals are $0, 1, 2, \dots, \omega, \omega+1, \dots$ where $\omega$ is the order type of $\mathbb{N}$, and $\omega+1$ is the order type of $\mathbb{N}$ together with a maximum element.
3
votes
Finite Number of Registers and Computable Well-Orderings
Unfortunately, I don't fully understand your question, concerning the registers.
But it seems that part of what is at stake is to find a computable relation on the natural numbers having order-type …
- 236k
1
vote
Accepted
Confusion regarding the requirements for a recursive ordinal notation
The subset of natural numbers referred to is essentially the set of notations, encoded as natural numbers. In your example, it is the set of numbers having form $2^n$ or $2^n\cdot 3$. The ordinal nota …
- 236k
8
votes
Accepted
Are two recursive well-orderings with the same order type recursively isomorphic?
The answer is no, not even for relations with order type $\omega$, and not even for primitive recursive relations.
To see this, let $\leq$ be the usual order relation on the natural numbers. And let …
- 236k
14
votes
Accepted
Cardinality of $\omega\uparrow^\omega\omega$
The Knuth arrow notation is most often defined only on the natural numbers, but the central idea of it
can be easily extended to the ordinals, for example as follows:
$$\alpha\uparrow^0\beta=\alpha\b …
- 236k
2
votes
Ordinal vs. cardinal dimension
One natural context where one thinks of the exponent or dimension in the space as having an order is with the function space $\kappa^\lambda$ considered under the relation of eventual domination. That …
- 236k
1
vote
A few questions about Tychonoff plank
To see that $S$ is countably compact, suppose that we have a countable open cover $\mathcal{U}$. Notice that $S$ is the union of the nested chain of subspaces
$$(\omega_1+1)\times (\alpha+1)\cup (\alp …
- 236k
6
votes
Accepted
Embedding large countable ordinals into the complex plane
Yes, in fact every countable ordinal embeds into the rational numbers in this way, an order-preserving map as a closed set of rational numbers.
Let me give three proofs.
First, one can easily prove th …
- 236k
4
votes
Accepted
An ordinal not $\Sigma_1$ stable in alpha must be in the hull of smaller parameters in $L_\a...
Since $\beta$ is not $\Sigma_1$-stable in $\alpha$, there is some new $\Sigma_1$ fact $\exists\xi\varphi(\xi,\vec a)$ true in $L_\alpha$ but not in $L_\beta$, for some $\vec a\in L_\beta$. It follows …
- 236k
7
votes
Who wins infinite Hex?
This doesn't answer the question that was asked, but rather an alternative kind of infinite Hex, played on the infinite hexagonal lattice board as shown below. I am posting it because people intereste …
- 236k
19
votes
What is the idea behind stationary sets?
One answer to your question about intuition is simply that stationary sets arise very naturally once you begin to think of the natural measure surrounding club sets. The stationary sets are simply tho …
- 236k
4
votes
Accepted
Restriction of a locally finitely supported function on an ordinal is finitely supported?
If the function has infinitely many nonzero values below some ordinal $\alpha$, let $\beta$ be the supremum of the locations of the first $\omega$ many instances. So $\beta\leq\alpha$ and every neighb …
- 236k
6
votes
First-order definable bijection between $P(On)$ (or $No$) and $V$? (Is this equivalent to $V...
$\newcommand\Ord{\text{Ord}}\newcommand\HOD{\text{HOD}}$Let me
point out that the existence of a definable bijection of $V$ with
$P(\Ord)$ is equivalent to the assertion that the universe is
Leibnizia …
- 236k
12
votes
Ramsey Theorem for the class ORD
Ali Enayat and I have proved that with respect to definable classes, Ord is NOT weakly compact. In particular, we show, in every model of ZFC,
there is a definable Ord-tree with no definable cofina …
- 236k
14
votes
Accepted
What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$?
Let me get things started with some simple observations.
Note that given any countable sequence of functions $f_n$, we can
by diagonalization construct a function eventually dominating all
of them, $ …
- 236k
8
votes
Accepted
Large cardinals disproving J=L
Unfortunately, your question seems to be based on a misunderstanding of the $J$ hierarchy. There is no such large cardinal hypothesis.
The reason is that already ZFC and indeed much weaker theories s …
- 236k