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Prove for each function $f:\mathbb Q\times\mathbb Q\to\mathbb R$ there exists a function $g:\mathbb Q\to\mathbb R$ such that $f(x,y)\le g(x)+g(y)\,\forall x,y\in\mathbb Q$.

Find a function $f:\mathbb R\times\mathbb R\to\mathbb R$ for which there is no function $g:\mathbb R\to\mathbb R$ such that $f (x,y)\le g (x)+g (y)\,\forall x,y\in\mathbb R $.

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    $\begingroup$ One should note that the problems are of purely settheoretic nature. In the first question $\mathbb{Q}$ can be replaced by $\mathbb{N}_0$ and then $g(x) := \max_{y,z \in \mathbb{N_0} :y+z \leq 2x} |f(x,y)|$ (of course very crude) does the job. For the second question I think $\mathbb{R}$ should be well ordered and then ? $\endgroup$
    – Dieter Kadelka
    Commented Jun 7, 2020 at 11:45
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    $\begingroup$ I am probably missing something, but why was the question closed? Is there an obvious answer to the second question? $\endgroup$
    – abx
    Commented Jun 8, 2020 at 3:47
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    $\begingroup$ Such questions are closed, though I was not involved, because there is no question. It sounds like a (hard) exercise for extra credits. $\endgroup$
    – András Bátkai
    Commented Jun 8, 2020 at 6:16
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    $\begingroup$ The second "question" was answered in a more general setting in the comments of this subsequent question where the exercise-style was not followed and that question was well-received. This makes it a duplicate, but rather than closing on this grounds, one could remove the second question and focus on the 1st one. But at the same time the first question is possibly too easy to deserve a separate question (it's claimed as easy in the linked question). $\endgroup$
    – YCor
    Commented Jun 8, 2020 at 8:35

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