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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
S Jun 2, 2014 at 11:23
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