Graph theory: Are infinite trees planar? I think countable trees are, but not uncountably infinite trees, apparently. How does one construct such a tree?
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Andrés E. Caicedo
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John Smith
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Draw the $x$-axis. Your vertices are the integers, and your edges are $(n,n+1)$ for all $n\in\mathbb{Z}$. – vadim123 Nov 23 '15 at 21:48
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Relevant: Can there exist an uncountable planar graph? – hmakholm left over Monica Nov 23 '15 at 21:52
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Oh wait I was confused. Is it just that there isn't enough "room" to put anything? So if there was a higher cardinality analogue of R^n then the graph might be "planar". – John Smith Nov 23 '15 at 22:00
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It is easy to construct a class of such graphs starting with a root and correspond the childs to an interval and so on - representation is some kind of Sierpinski triangle. But going to 2nd step childs things gets more complicated. – z100 Nov 23 '15 at 22:20
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So if we have an infinite tree the cardinality of the continuum (vertex set)... – John Smith Nov 23 '15 at 22:30
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1If you have a tree with a vertex set big enough (for example $2^{\mathfrak{c}}$), then it cannot be injected in the plane. – Crostul Nov 23 '15 at 22:36
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Yes, intuitively any higher cardinality tree cannot be embedded into plane, I don't see how to construct something like that. – z100 Nov 23 '15 at 22:39