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The question is as follows:

Let $A$ be a subset of $\mathbb{R}^2$. A section of $A$ is a subset of $\mathbb{R}$ of the form $\{t \in \mathbb{R} : a + tb \in A\}$, where $a,b \in \mathbb{R}^2$ with $b \neq 0$. Prove that there does not exist a set $A \subset \mathbb{R}^2$ such that every set $S \subset \mathbb{R}$ is a section of $A$.

I think I'm misunderstanding this problem. If $A$ contains the line $y=x$, then any $t \in \mathbb{R}$ lies in the section $\{t \in \mathbb{R}: (0,0)+t(1,0) \in A\}$? Or must it be the case that, if $S$ corresponds to some section, it must contain all values of $t$ that satisfy the property?

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    I think you have a typo in your title which indicates a misunderstanding: The question asks whether every *subset* of $\mathbb R$ could be a section, not whether every *element* of $\mathbb R$ is in a section (which is what you've shown). Does that clarify things for you? – 2017-01-12
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    Your example shows that if $A$ contains the diagonal line, then $\mathscr{R}$ is a section of $A$. That’s all that it shows. – 2017-01-12
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    No, I understood that. But there was a misunderstansing indeed, I was interpreting it as every subset is a subset of some section, which is silly. – 2017-01-12

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HINT: Each section of $A$ is determined by an ordered pair $\langle a,b\rangle$ of points of $\Bbb R^2$.

  • What is the cardinality of $\Bbb R^2\times\Bbb R^2$? There are at most that many different sections of any given $A\subseteq\Bbb R^2$.

  • What is the cardinality of $\wp(\Bbb R)$?

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    Yes, I see now. The cardinality of that is R. We cannot inject the powerset of R into the sections. However - must I show that every subset, if we assume it to be affiliated with a section, is only affiliated with ONE ordered pair? I can see that if two subsets belong to the same section, they are the same. But must I necessarily show that every set is at most one section – 2017-01-12
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    @JP.: It’s entirely possible for a section of $A$ to be generated by many different pairs $\langle a,b\rangle$. It’s enough to observe that there are simply too many subsets of $\Bbb R$ for all of them to be sections even if each ordered pair produced a different section. – 2017-01-12
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    You mean, if each ordered pair didn't. Alright, thank you, I understand now. – 2017-01-12
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    @JP.: No, I mean even if each ordered pair *did* produce a different section: that’s the situation that maximizes the number of sections, and even that isn’t enough. \\ You’re welcome. – 2017-01-12
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    However, what about this follow-up question? Is there a set A such that each countable subset of R is a section of A. – 2017-01-12
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    @JP.: Yes, there is. HINT: There are $\mathfrak{c}$ countable subsets of $\Bbb R$ and the same number of horizontal lines in the plane. – 2017-01-12
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    Ah yeah. Is there a straightforward way to show that there are that many countable subsets of $\mathbb{R}$? – 2017-01-12
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    @JP.: There are $\mathfrak{c}^\omega=\left(2^\omega\right)^\omega=2^{\omega\cdot\omega}=2^\omega=\mathfrak{c}$ countable subsets of $\Bbb R$. – 2017-01-12