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31 votes
Accepted

Characterizing $\mathbf{R}$ as an ordered group

The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following ...
Jochen Glueck's user avatar
22 votes
Accepted

Suprema of directed sets

Yes, a poset that has suprema of all chains also has suprema of all directed sets. This is known, and I vaguely recall seeing it attributed to Solovay. The proof consists of showing, by induction on ...
Andreas Blass's user avatar
12 votes
Accepted

Associative mean

Yes there are more such functions, even if we require them to be symmetric, monotonic and continuous. For example pick any $C$ and take $m_C(a,b)=\min(a,b)$ if $a,b\ge C$, $m_C(a,b)=\max(a,b)$ if $a,...
Yaakov Baruch's user avatar
12 votes

Associative mean

Let $(L, <)$ be any linear order. Fix an arbitrary well-order $\sqsubset$ on $L$. For any $a\le b$ define $m(a,b)=m(b,a):= $ the $\sqsubset$-least element in the interval $[a,b]$. Then $m$ is ...
Goldstern's user avatar
  • 14k
9 votes
Accepted

Upward Löwenheim–Skolem theorem for well-ordered models with/without measurable cardinals

A very nice collection of questions. Here are a few things one can say to get started. If $\kappa$ is a measurable cardinal and $T$ has a well-ordered model of size at least $\kappa$, then it has ...
Joel David Hamkins's user avatar
6 votes

Classifying the endofunctors of the category $\Delta$ of finite linear orders

See Edgewise subdivision and simple maps by Knut Berg (supervised by me), Generalized edgewise subdivisions by Katerina Velcheva (supervised by Clark Barwick) and the earlier MathOverflow question ...
John Rognes's user avatar
  • 9,263
6 votes
Accepted

Is there an explicit linear extension for the subsequence partial order?

Note that two words of the same lengths are comparable if and only if they are equal. So you can order the words in the following way: $$X\prec^* Y\iff |X|<|Y|\text{ or } (X<_{\rm Lex}Y \text{ ...
Asaf Karagila's user avatar
  • 39.7k
6 votes
Accepted

Ordinal notations within non-standard models of arithmetic

I'll provide a description of $(\alpha)^{\mathfrak{A}}$ in the case of countable $\mathfrak{A}$. I base my answer on observations by Emil Jeřábek (see discussion below the initial question), but of ...
Fedor Pakhomov's user avatar
6 votes

Characterization of Archimedean linearly ordered monoids

By a classic result of Alimov (1950), a fully ordered semigroup $S$ is order isomorphic to a subsemigroup of the additive group of real numbers if and only if it is cancellative and contains no ...
Philip Ehrlich's user avatar
5 votes
Accepted

Is there literature on finite geometries with ordered lines?

Yes, this has been studied and is indeed known as ordered geometry or the study of betweenness spaces: https://en.m.wikipedia.org/wiki/Ordered_geometry
Bjørn Kjos-Hanssen's user avatar
5 votes
Accepted

Fundamental theorem of linear orders

As already observed, you may as well assume $(\Omega,\leq)$ is $(\mathbb{Q},\leq)$. In this case, of course one condition on $\Omega_1$ and $\Omega_2$ ensuring your desired property is finiteness. ...
Gabe Conant's user avatar
  • 3,274
4 votes

Ordinal notations within non-standard models of arithmetic

I spent some time thinking about order-types of models of arithmetic, and of order-types of structures they interpret (my PhD). I think there are several formulations of your question: (1) about ...
andrey bovykin's user avatar
4 votes

Reference for tree of bad sequences of WPO

OK I found an answer myself. Blass and Gurevich (2008) fill in the missing details (and reprove the result in its entirety). Andreas Blass and Yuri Gurevich, Program termination and well partial ...
Gabriel Nivasch's user avatar
4 votes

Is a Banach lattice isomorphic to a Hilbert space in fact a Hilbert lattice?

Yes. This is a consequence of the fact that semi normalized unconditionally basic sequences in a Hilbert space are Riesz bases. For a theorem quoting proof of what you want, note that that the lattice ...
Bill Johnson's user avatar
  • 31.5k
3 votes
Accepted

A closed subset of a Dedekind-complete order has subspace topology equal to order topology

While I agree that it's pretty direct to show, I was unable to find a reference for a proof of this fact myself (I thought it was in Willard, but I thumbed through my copy and failed to find it). So ...
Steven Clontz's user avatar
3 votes

Ordering preference for two zero mean Gaussian outcomes

It seems like it is high time to handle this one. The main difficulty here is that there seems to be no conceptual reason for the inequality to be true: it just comes up valid before one numerical ...
fedja's user avatar
  • 61.9k
3 votes

Topology on set of "real lower bounds"

A slight modification of this space, where you let $t$ range over $[0,1]$ instead of $\mathbb R$, is known as the double arrow space or the split interval. You can learn more about it here or here.
Will Brian's user avatar
  • 18.5k
3 votes
Accepted

Permutations which respect a partial order

In more common language, the question is whether for each finite poset, any two linear extensions of the poset are related by a series of adjacent transposition that exchange incomparable elements. ...
Sam Hopkins's user avatar
  • 24.2k
3 votes

Self-embeddings of uncountable total orders

I may be missing something, but it seems to me that the exact same statement holds for all infinite total orders. Indeed, if $\mathbb Q$ does not embed in $\Omega$, then $\Omega$ is scattered, hence ...
Emil Jeřábek's user avatar
3 votes
Accepted

A strictly decreasing function between uncountable subsets of the reals

In Todorcevic's book "Partition Problems in Topology" I have found Proposition 8.4(c) saying that under OCA for any uncountable sets $X,Y$ of reals there exists a strictly increasing function $f:Z\to ...
Taras Banakh's user avatar
  • 41.8k
3 votes
Accepted

Name for this algebraic structure?

As Emil Jeřábek correctly suggests in the comments above, $\mathbb{M}$ is exactly the non-negative part of a discretely ordered group -- I humbly offer a proof to close the thread. The proof that the ...
Alec Rhea's user avatar
  • 10.1k
3 votes

Associative mean

Let us, indeed, describe "monotone, continuous, associative means" on $[0,1]^2$. I'll just write $a*b$ instead of $m(a,b)$ to make long expressions easier to comprehend. Let $c=0*1,d=1*0$. Then, for ...
fedja's user avatar
  • 61.9k
2 votes

Order homomorphism functions on $\omega_1$

The answer to the above question is no (there is no such mapping $\psi$). The question asked above is an order-theoretic restatement of a question from general topology: Question. Does $\omega_1$ (...
Mirko's user avatar
  • 1,375
2 votes
Accepted

Kruskal's tree theorem and $\Pi_1$ sentences of linear orderings with finitely many constants

Towards a contradiction, suppose not. Then we can find a sequence $\phi_0, \phi_1, \dots$ from $\Phi$ such that for all $i$, $LO, \phi_0, \dots, \phi_i \not \vdash \phi_{i+1}$. So for each $i$, ...
Dan Turetsky's user avatar
  • 3,073
1 vote
Accepted

Find an order-embedding of $S_3\times{\bf2}\times{\bf2}$ into ${\mathbb Z}^4$

We present an embedding into $\mathbb Z^3$. It is symmetric with respect to the premutation of the coordinates in $S_3$, so it suffices to show the images of the following elements $((x,y,z),a,b)\in ...
Ilya Bogdanov's user avatar
1 vote
Accepted

Order-embeddability of ${\frak b}$ and ${\frak d}$ in $\mathbb{R}$

Assume that $i : \alpha \to \mathbb{R}$ is an order-embedding of some ordinal $\alpha$ into $(\mathbb{R},<)$. We can modify $i$ to yield an order embedding $j : \alpha \to \mathbb{Q}$ by induction ...
Arno's user avatar
  • 4,667
1 vote

What is the dimension of a subspace of the product of $n$ linearly ordered compacta

In an old PhD-thesis "Finite products of locally compact ordered spaces" by J. van Dalen (Vrije Universiteit, Amsterdam) from 1972, I found (I could not find a paper with the result, so far, as I have ...
Henno Brandsma's user avatar
1 vote

Order homomorphism functions on $\omega_1$

I have been thinking about this question since it was bumped up. I am burned out now, so I thought I would post a few basic things that I noted. First, as you noted that any function $f \in \mathcal ...
SSequence's user avatar
  • 881
1 vote

An algebraically generated set of linear orders

Thanks to Goldstern for the reference. Very helpful! I finally found time to look through a copy of "Linear Orderings" by Joseph Rosenstein, which is a very nice book. It turns out that Rosenstein ...
Jeremy Brazas's user avatar

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