Timeline for Equality-preserving embeddings of finite trees
Current License: CC BY-SA 3.0
15 events
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Jun 30, 2014 at 22:20 | vote | accept | M Carl | ||
Jun 27, 2014 at 20:04 | answer | added | Tony Huynh | timeline score: 3 | |
S Jun 2, 2014 at 11:23 | history | suggested | F. C. |
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Jun 2, 2014 at 11:19 | review | Suggested edits | |||
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Jun 2, 2014 at 10:03 | answer | added | GabrielG | timeline score: 4 | |
Jun 1, 2014 at 18:35 | history | edited | M Carl | CC BY-SA 3.0 |
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May 31, 2014 at 23:28 | history | edited | M Carl | CC BY-SA 3.0 |
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May 31, 2014 at 23:21 | comment | added | M Carl | Yes. I think that is the way the term "embedding" is usually understood in this context. | |
May 31, 2014 at 23:01 | comment | added | bof | By "embedding in the usual graph-theoretical sense" I guess you mean homeomorphic (rather than, say, homomorpic) embedding. Did I guess right? | |
May 31, 2014 at 14:46 | comment | added | Emil Jeřábek | The label-free formulation of the question is that you consider finite trees endowed with an equivalence relation, and you quasiorder them by tree embeddings that are homomorphisms for the equivalence relations. | |
May 31, 2014 at 8:54 | history | edited | M Carl | CC BY-SA 3.0 |
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May 30, 2014 at 8:31 | comment | added | M Carl | Sure. But here we have all natural numbers available as our colours, not just a finite set; so while each tree is of course coloured by finitely many colours only, all the trees together may have infinitely many different colours, so Kruskal (or Nash-Williams, for bqo) is not applicable (at least not obviously, or at least not obviously for me). | |
May 29, 2014 at 14:48 | comment | added | Tobias Schlemmer | I don't understand your question. Isn't it true that every finite subset of the natural numbers is a special case of a finite set? | |
May 27, 2014 at 16:59 | review | First posts | |||
May 27, 2014 at 17:03 | |||||
May 27, 2014 at 16:43 | history | asked | M Carl | CC BY-SA 3.0 |