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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 continous. 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](http://books.google.com/books?id=8EWnEh18rVgC&pg=PA14).2012-04-13
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    Did you replace the question? Please don't do that. Ask a new question instead.2012-04-13
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    oops i am sorry!!2012-04-13
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    oops! I made a mistake, I am so sorry! thank you.2012-04-13
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    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.2012-04-13

2 Answers 2

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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?2012-04-13
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    It is $\mathbb Q$-linear, but you have to prove it!2012-04-13
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    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 $(0,0)$.2012-04-13
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    The closed graph theorem doesn't imply that $\Gamma$ is not closed. You need linearity for this?2012-04-13
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    @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.2012-04-13
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    @t.b.: Thanks for the info. Good to be aware of.2012-04-13
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    Indeed, that was the point really (but I should have checked that the Wikipedia page actually mentioned that version!) In any case, one does not need the graph to be non-closed, so I simply removed the reference to the CGT)2012-04-13
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    @MarianoSuárez-Alvarez: Utilizing the linearly independent points is very slick. My functional analysis is very much at the introductory level.2012-04-13
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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?2012-04-13
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    would you please elaborate?2012-04-13
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    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 $\mathbb{R}$.2012-04-13
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    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.) http://groups.google.com/group/sci.math/msg/98d0bb02228bd4bd and http://groups.google.com/group/sci.math/msg/4016347301a711402012-04-13
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    Not that it matters, but why unaccept the answer after, what, almost 8 years?2018-02-14