$f:\mathbb{R}\rightarrow\mathbb{R}$ 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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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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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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