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If not, how many points can be guaranteed?

Also, I'm not sure about my tag. This is a pretty general question. I figured General Topology is close.

EDIT: Someone paraphrased this nicely. "Let $S\subset [0,1]^2$ be uncountable. Is there a line which contains infinitely many points of $S$?"

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    $\begingroup$ Any line passing through the unit square will contain uncountably many points in the unit square. Think about the $x$-axis. It passes through $(0,0)$ to $(1,0)$ so every point $(a,0)$ with $0\leq a\leq 1$ will be in the unit square. There are uncountably many $0\leq a\leq1$. $\endgroup$
    – Jam
    Sep 30, 2017 at 22:30
  • $\begingroup$ Any line of non-zero length intersecting the unit square in more than a single point will have uncountably many point in common with the unit square. This is esentially because both the sets (the square and the line) are continuous. $\endgroup$ Sep 30, 2017 at 22:32
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    $\begingroup$ I think the question is "Let $S \subset [0,1]^2$ be uncountable. Is there a line which contains infinitely many points of $S$?" $\endgroup$ Sep 30, 2017 at 22:36
  • $\begingroup$ Yes, Patrick. That's what I meant. $\endgroup$
    – Zohee
    Sep 30, 2017 at 22:37
  • $\begingroup$ Alternatively, can you find an uncountable $S \subset [0,1]^2$ such that all lines that intersect $S$ intersect in finitely many points? $\endgroup$
    – Rustyn
    Sep 30, 2017 at 22:42

2 Answers 2

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Hint: Think about a cirlce $S$ inside $[0,1]\times [0,1].$

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  • $\begingroup$ Dang, beat me to it! +1. $\endgroup$ Sep 30, 2017 at 22:45
  • $\begingroup$ Very nice. I suppose most (continuous) natural functions which aren't actually linear in part will work. Certainly all non-linear polynomials, the exponential, logarithm, rational functions, just about anything I can think of. I'm putting in continuous since you can always put in countably many discontinuities to create a line full of holes. $\endgroup$
    – DRF
    Oct 1, 2017 at 1:29
  • $\begingroup$ @DRF Yes, the graph of any nonlinear real analytic function on $\mathbb R$ that intersects $(0,1)\times (0,1)$ will do. $\endgroup$
    – zhw.
    Oct 1, 2017 at 1:57
  • $\begingroup$ @NoahSchweber Looks like I was typing an answer while you made the identical point in a comment. I owe you 10 up votes. $\endgroup$
    – zhw.
    Oct 1, 2017 at 1:59
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There is a subset $S$ of size $\omega_1$ of the unit square such that any line contains at most two points of $S$. Start by picking two points of the square. Draw the line through them. Pick a third point, avoiding the line you drew. Draw the lines from the new point through all the previous points. Pick the next point avoiding all the lines you drew. Continue by transfinite induction until you have $\omega_1$ points. At all stages of the construction you have only used a countable number of points, so there are a countable number of lines drawn. As Lebesgue measure is countably additive, the area of all the lines is zero so there are points available to continue.

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  • $\begingroup$ You beat me to it - I was tediously writing out the inductive definitions of each set :) $\endgroup$ Sep 30, 2017 at 22:44
  • $\begingroup$ Or draw a circle. $\endgroup$ Sep 30, 2017 at 22:44
  • $\begingroup$ @NoahSchweber Yes soo much easier but I also started with assuming CH and doing an $\omega_1$ length diagonalization. Wonder how many of us there were. $\endgroup$
    – DRF
    Oct 1, 2017 at 1:15

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