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A question I found hard to solve.

Prove that a countable set of parabolas $\alpha (y-\beta)^2+\gamma=x$, for $\alpha, \beta, \gamma \in \Bbb R$, doesn't cover the entire $xy$ plane

Thanks in advnace for any assistance.

Did
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henry
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    The union of all such parabolas is in fact $\mathbb{R}^2$, since $(x,y)$ solves $\alpha(y-\beta)^2+\gamma=x$ for $\alpha=\beta=0$ and $\gamma=x$. – Abel Apr 29 '13 at 10:02
  • Even considering only real parabolas (and not degenerated ones like $x=\gamma$), i.e. $\alpha\neq 0$, the plane $\mathbb{R}^2$ is fully covered. – saposcat Apr 29 '13 at 10:11
  • maybe you mean a non-intersecting set of parabola with $\alpha\neq 0$ ? – saposcat Apr 29 '13 at 10:12

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Let $P = \{(\alpha,\beta,\gamma)\} \subseteq \mathbb{R}^3$ be countable and fix some $y\in\mathbb{R}$. Note that for any triple $(\alpha,\beta,\gamma)\in P$ there is precisely one $x$ solving $\alpha(y-\beta)^2+\gamma=x$. Hence the union of all $(x,y)$ such that $x$ solves $\alpha(y-\beta)^2+\gamma=x$ for some $(\alpha,\beta,\gamma)\in P$ is countable.

Thus the union of countably many such parabolas intersects any vertical line only countably many times and hence does not cover $\mathbb{R}^2$.

Abel
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