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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 $S$?"

zhw.
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user402313
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    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 $0\leq a\leq1$. – Jam Sep 30 '17 at 22:30
  • 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. – Bobson Dugnutt Sep 30 '17 at 22:32
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    I think the question is "Let $S \subset [0,1]^2$ be uncountable. Is there a line which contains infinitely many points of $S$?" – Patrick Stevens Sep 30 '17 at 22:36
  • Yes, Patrick. That's what I meant. – user402313 Sep 30 '17 at 22:37
  • Alternatively, can you find an uncountable $S \subset [0,1]^2$ such that all lines that intersect $S$ intersect in finitely many points? – Rustyn Sep 30 '17 at 22:42
  • @RossMillikan: Yes, but it was 3 seconds too late, so I deleted it. – Moishe Kohan Sep 30 '17 at 23:13
  • @PatrickStevens Your rephrasing is now the subject title. – zhw. Oct 01 '17 at 01:43

2 Answers2

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

zhw.
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  • Dang, beat me to it! +1. – Noah Schweber Sep 30 '17 at 22:45
  • 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. – DRF Oct 01 '17 at 01:29
  • @DRF Yes, the graph of any nonlinear real analytic function on $\mathbb R$ that intersects $(0,1)\times (0,1)$ will do. – zhw. Oct 01 '17 at 01:57
  • @NoahSchweber Looks like I was typing an answer while you made the identical point in a comment. I owe you 10 up votes. – zhw. Oct 01 '17 at 01:59
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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.

Ross Millikan
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