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Apr 22, 2019 at 20:55 review Close votes
Apr 23, 2019 at 5:20
Apr 22, 2019 at 13:43 vote accept Tim Campion
Apr 21, 2019 at 20:06 answer added Trevor Wilson timeline score: 14
Aug 29, 2018 at 14:44 comment added Tim Campion The construction of an embedding $Ord \to C$ given above only works in certain $C$, such as the category of graphs, but that's fine.
Aug 29, 2018 at 14:42 comment added Tim Campion @HarryGindi Not only is not every linear a well-order, but the morphisms of well-orderings are also much more restrictive than those of linear orders. For the statement, see Lemma 6.2 of Adamek-Rosicky. From a large discrete full subcategory $(A_i)_{i \in Ord} \subseteq C$, one constructs an embedding $Ord \to C$ by sending $i \mapsto \amalg_{j < i} A_j$. Conversely, an embedding $Ord \to C$ for $C$ locally presentable contradicts the form of Vopenka's principle which says that any class $(A_i)_{i \in Ord} \subseteq C$ admits a morphism $A_i \to A_j$ for some $i < j$.
Aug 29, 2018 at 14:34 vote accept Tim Campion
Apr 22, 2019 at 13:43
Aug 29, 2018 at 13:00 comment added Alex Kruckman @HarryGindi Not every linear order is a well-order.
Aug 29, 2018 at 10:01 comment added Harry Gindi I am trying to recall this statement of VP (that ORD admits no full embedding in any locally presentable cat) and I can't find it. Isn't ORD the skeleton of LinOrdSet, though?
Aug 29, 2018 at 8:55 answer added Ivan Di Liberti timeline score: 5
Aug 29, 2018 at 2:18 history edited Tim Campion CC BY-SA 4.0
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Aug 29, 2018 at 1:58 history asked Tim Campion CC BY-SA 4.0