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I have some difficulties in the following question.

Let $S$ be a dense subset in $\mathbb{R}^n$. Can we find a straight line $L\subset\mathbb{R}^n$ such that $S\cap L$ is a dense subset of $L$.

Note. From the couterexample of Brian M. Scott, I would like to ask more. If we suppose that $S$ has a full of measure. Could we find a straight line $L$ such that $S\cap L$ is a dense subset of $L$?

blindman
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  • Pick, for example $S = \mathbb{Q}\times\mathbb{R}$. Then, $S$ is a dense subset of $\mathbb{R}^2$. On the other hand, for the line $L = {\pi}\times\mathbb{R}$, we have $S\cap L = \emptyset$. – Lord Soth Apr 08 '13 at 23:31
  • @LordSoth But that doesn't answer the question. – Harald Hanche-Olsen Apr 08 '13 at 23:33
  • @HaraldHanche-Olsen Sorry, I read the question as "whether we can do this for any dense subset of $\mathbb{R}^n$." I guess the OP asks whether we can construct an $S$ such that we can find an $L$ with that property? – Lord Soth Apr 08 '13 at 23:35
  • @LordSoth: The question is whether every dense subset of $\Bbb R^n$ has dense intersection with some straight line in $\Bbb R^n$; the answer is no. – Brian M. Scott Apr 08 '13 at 23:37
  • @BrianM.Scott Oh well, thanks. I need some rest... – Lord Soth Apr 08 '13 at 23:42

1 Answers1

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For each $n>1$ it’s possible to construct a dense subset $D$ of $\Bbb R^n$ such that every straight line in $\Bbb R^n$ intersects $D$ in at most two points.

Let $\mathscr{B}=\{B_n:n\in\omega\}$ be a countable base for $\Bbb R^n$. Construct a set $D=\{x_n:n\in\omega\}\subseteq\Bbb R^n$ recursively as follows. Given $n\in\omega$ and the points $x_k$ for $k<n$, no three of which are collinear, observe that there are only finitely many straight lines containing two points of $\{x_k:k<n\}$; let their union be $L_n$. $L_n\cup\{x_k:k<n\}$ is a closed, nowhere dense set in $\Bbb R^n$, so it does not contain the open set $B_n$, and we may choose $x_n\in B_n\setminus\big(L_n\cup\{x_k:k<n\}\big)$.

Clearly $D$, so constructed, is dense in $\Bbb R^n$, since it meets every member of the base $\mathscr{B}$, and by construction no three points of $D$ are collinear.

Brian M. Scott
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  • The question gives $S$ as if it is arbitrary, then you try to find the line $L$. He is not asking to construct the dense set, he is asking to find a specific line for a given dense set. – Integral Apr 08 '13 at 23:41
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    @Integral: The point is that if you can construct a dense set that does not have dense intersection with any straight line, then the answer to the question is no: given an arbitrary dense subset of $\Bbb R^n$ for $n>1$, it is not necessarily possible to find such a line $L$. This, this construction does answer the question. – Brian M. Scott Apr 08 '13 at 23:43
  • Ok, I agree. But you constructing a set $D$ as you did proves that is impossible to construct another dense set that does not have dense intersection with any straight line? – Integral Apr 08 '13 at 23:46
  • naah...never mind, I get it now. – Integral Apr 08 '13 at 23:48
  • I misundertood what you did. – Integral Apr 08 '13 at 23:48
  • @Brian M. Scott: Thank you for your nice answer. If we suppose that $S$ has a full of measure. Could we find a straight line $L$ such that $S\cap L$ is a dense subset of $L$? – blindman Apr 08 '13 at 23:53
  • @blindman: You’re welcome. Off the top of my head I don’t know; it wouldn’t really surprise me either way. – Brian M. Scott Apr 08 '13 at 23:55
  • @Brian M. Scott: Dear Sir. I have an interesting question in the following link http://math.stackexchange.com/questions/346937/on-the-weak-and-strong-convergence-of-an-iterative-sequence I woulk to to ask your kind comments? – blindman Apr 12 '13 at 02:21