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There is a related usage in set theory in the context of Fodor's lemma, where we often consider functions $f$ on the ordinals with the property that $f(\alpha)<\alpha$. These are called the regressive …

I don't see any reason not to use the same terminology as one uses in the case of a partial order (or a pre-order), since a set $S$ is closed in your sense with respect a relation if and only if it is …

The lexical order of a well ordered list of orders $\langle X_\alpha,\leq_\alpha\rangle$ is the order on the product space $\Pi_\alpha X_\alpha$ placing $s$ before $t$ if $s_\alpha\lt_\alpha t_\alpha$ …

Let me address merely the suggestion the OP makes in the comment: whether this ordinal can be specified from the cofinality of the field. The answer is no, because any ordered field $F$ can be eleme …

A model $M$ of a theory $T$ is existentially closed with respect to that theory, if for any quantifier-free formula $\varphi$ and any objects $\vec a$ in $M$, if there is model $N$ of the theory $T$ e …

No, clearly not, because you could put junk on top. But even if you avoid this by insisting that the map is surjective, there are counterexamples. Consider the map of Eric Wofsey's recent answer, wh …

The answer is yes, because any two nonempty open sets have points in common. And this also shows directly that the space is not Hausdorff. From what you describe in the other question, the topology i …

$\newcommand{\P}{\mathbb{P}} \newcommand{\L}{\mathbb{L}} \newcommand{\Z}{\mathbb{Z}}$ The answer is yes, and you don't need the locally finite hypothesis. Also, you may assume that the embedding is …

One natural example would be the complete bipartite digraphs, which are all semi-transitive but (if nontrivial) are not transitive. A complete bipartite digraph is a digraph whose vertices partition …

If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of count …

Such a function can have only countably many points of discontinuity, since any discontinuity will be a jump discontinuity and hence the range will skip over an interval unique to that point, and so w …

If the ambient order is a lattice, they your order is indeed a lattice order. To see this, suppose that we have intervals (a,b) and (a',b'), in your sense that $a\le b$ and $a'\le b'$. If $b\lt b'$, t …

If $R$ and $S$ are binary relations, then the composition relation $R\circ S$ is usually defined by $a\mathrel{R\circ S} c$ if and only if there is $b$ such that $a\mathrel{R}b$ and $b\mathrel{S}c$. …

Indeed, your conjecture is correct. Theorem. If L is a complete lattice and S is a subset of L, then S is chain-closed iff S is filter-closed. Proof. Clearly filter-closed implies chain-closed, sin …

If one replaces the cofinality $\omega$ requirement by an arbitrary cofinality (and if one also insists that $\alpha_m\lt\alpha$, as seems intended), then the construction is the same as that of the C …