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Let $\kappa$ be an infinite cardinal and suppose $$n, d: \kappa \to \big((\kappa+1)\setminus \{0\}\big) = \{1, \ldots, \kappa\}$$ are arbitrary functions.

Is there $E \subseteq \big\{\{x,y\}: x\neq y \in \kappa\big\}$ such that the graph $G=(\kappa,E)$ has the following property?

For all $k\in \kappa$ there are exactly $n(k)$ elements of $\kappa$ that have degree $d(k)$.

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  • $\begingroup$ Is $\kappa$ countable? Title and anything else suggest that yes. $\endgroup$
    – Fedor Petrov
    Commented Feb 26, 2016 at 11:25
  • $\begingroup$ Oh sorry - have to change title! $\endgroup$
    – Dominic van der Zypen
    Commented Feb 26, 2016 at 13:35

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You might find the following paper useful: Degree sequences of infinite graphs, by Andreas Blass and Frank Harary.

The degree sequences of finite graphs, finite connected graphs, finite trees and finite forests have all been characterized. Our present purpose is to provide such characterizations in the infinite case.

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  • $\begingroup$ I can't view the article, so I gather you are saying that the answer is "No" to my question? $\endgroup$
    – Dominic van der Zypen
    Commented Feb 26, 2016 at 13:37
  • $\begingroup$ If my quick translation between your question and the notation from the paper is correct, the answer is in fact yes, and the graph can be taken to be a forest. $\endgroup$
    – Ben Barber
    Commented Feb 26, 2016 at 13:58

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