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$f:\mathbb{R}\rightarrow\mathbb{R}$ is a function such that for all $x,y$ in $\mathbb{R}$, $f(x+y)=f(x)+f(y)$. If $f$ is cont, then of course it has to be linear. But here $f$ is NOT cont. Then show that the set $\{{(x,f(x)) : x {\rm\ in\ } \mathbb{R}\}}$ is dense in $\mathbb{R}^2$.

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A proof is given e.g. in Functional equations in several variables By J. Aczél, Jean G. Dhombres p.14. – Martin Sleziak Apr 13 '12 at 7:29
thank you dear sir. – Un Chien Andalou Apr 13 '12 at 8:43
Did you replace the question? Please don't do that. Ask a new question instead. – Asaf Karagila Apr 13 '12 at 9:49
oops i am sorry!! – Un Chien Andalou Apr 13 '12 at 9:54
Don't just add a new, unrelated question to this question. Ask a new, separate question by clicking "ask question" at the top of the page. – Chris Eagle Apr 13 '12 at 10:06
up vote 2 down vote accepted

Let $\Gamma = \{ (x,f(x)) \}_{x \in \mathbb{R}}$. First show that the set $\Delta = \{ x | f(x) \neq 0 \}$ is dense in $\mathbb{R}$. Then show that $f$ is discontinuous at $0$, and that this implies that the closure of $\Gamma$ contains $\{0\}\times \mathbb{R}$. Then show that the closure of $\Gamma$ contains $\{x\}\times \mathbb{R}$, $\forall x \in \Delta$. Presumably the result will be obvious at this point.

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as we know if f is discontinous the kernel f must be dense, so how could it be capital delta dense? – Un Chien Andalou Apr 13 '12 at 6:49
would you please elaborate? – Un Chien Andalou Apr 13 '12 at 6:51
If $f(x_0) \neq 0$ for some $x_0$, then since $f(q x) = q f(x)$, $\forall q \in \mathbb{Q}$, clearly $\Delta$ is dense in $\mathbb{R}$. – copper.hat Apr 13 '12 at 7:04
A proof with some more details is given at the first URL. (The second URL contains a minor correction to something else in the first post.) and – Dave L. Renfro Apr 13 '12 at 15:19

Let $\Gamma$ be the graph.

If $\Gamma$ is contained in a $1$-dimensional subspace of $\mathbb R^2$, then it in fact coincides with that line. Indeed, the line will necessarily be $L=\{(\lambda,\lambda f(1)):\lambda\in\mathbb R\}$, and for all $x\in\mathbb R$ the line $L$ contains exactly one element whose first coordinate is $x$, so that $\Gamma=L$. This is impossible, because it clearly implies that $f$ is continuous.

We thus see that $\Gamma$ contains two points of $\mathbb R^2$ which are linearly independent over $\mathbb R$, call them $u$ and $v$.

Since $\Gamma$ is a $\mathbb Q$-subvector space of $\mathbb R^2$, it contains the set $\{au+bv:a,b\in\mathbb Q\}$, and it is obvious that this is dense in the plane.

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Is our f linear? – Un Chien Andalou Apr 13 '12 at 7:04
It is $\mathbb Q$-linear, but you have to prove it! – Mariano Suárez-Alvarez Apr 13 '12 at 7:05
Slick proof. You don't need closure to show the existence of two l.i. points. Discontinuity alone ensures that the graph cannot be contained in a line through $(0,0)$. – copper.hat Apr 13 '12 at 7:15
The closed graph theorem doesn't imply that $\Gamma$ is not closed. You need linearity for this? – copper.hat Apr 13 '12 at 7:26
@copper.hat: Banach proved that the open mapping theorem and the closed graph theorem also hold for Polish groups (second countable and metrizable with a complete metric) and homomorphisms, so linearity could be dispensed with, but of course it is serious overkill for the present question. – t.b. Apr 13 '12 at 8:43

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