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Is a dense subset of the plane always dense in some line segment?

The answer by André Nicolas presents a good and simple method of the existence of a dense subset with no 3 points in collinear. But according to what I can understand, the proof only guarantees points $a_n\in B(1/n, q_i)$ to be not in collinear with other points $a_m\in B(1/m, q_i)$ for the same centre point $q_i$.

So how to assert that points $a\in B(1/m, q_i), b\in B(1/k, q_s)$ and $c\in B(1/j, q_t)$ (i.e. points with distinct centre points) are still not in collinear with each other, or I am missing some crucial points in that proof which solves such issue?

Thanks!

Community
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Toad Jiang
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1 Answers1

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The explicit construction in Ilmari Karonen's answer to the linked question ensures the following (starting from any enumerated dense subset $Q=\{q_1,q_2,\ldots\}$ of $\Bbb R^2$):

  • A point $a_n$ is picked from each of the balls $B(\frac1n,q_n)$
  • $a_n$ avoids the finitely many lines $a_ia_j$ with $1\le i<j<n$

The second property guarantees that the set $A=\{a_1,a_2,a_3,\ldots\}$ does not contain three collinear points. If $a_i,a_j,a_n$ were such collinear points (where wlog. $i<j<n$) then apparently we did not obey the second property when choosing $a_n$.

The first property guarantees that $A$ is dense in $\Bbb R^2$: For any $x\in\Bbb R^2$ we find a sub-sequence of $Q$ converging to $x$. Then the corresponding subsequence of $A$ also converges to $x$.

Hagen von Eitzen
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