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If we say that $\mathcal{S}$ is a family of disjoint triangles on a plane, is $\mathcal{S}$ a countable or an uncountable set?

I believe that it is an uncountable set since $\mathbb{R}$ is uncountable and our triangles' corner coordinates are $\in \mathbb{R}$, but I cannot prove this formally, any tip would help.

Bekir Kırkıcı
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  • We have no information on $\mathcal S$, only that it is "a family of..." - so it can also be the empty set. – dan_fulea Sep 06 '19 at 00:28
  • $\mathcal S$ could be countable or finite : I mean, a set consisting of just one triangle satisfies the conditions that $S$ does. However, $\mathcal S$ can't be uncountable because there's a point with rational coordinates trapped inside each triangle, and there are only countably many such coordinates in $\mathbb R^2$. – Sarvesh Ravichandran Iyer Sep 06 '19 at 00:29
  • Please define the notion of a "triangle" exactly. – dan_fulea Sep 06 '19 at 00:31

1 Answers1

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There is a point with rational coordinates in the interior of each triangle. Any such point can only be in one of the triangles, since the triangles are disjoint. There are only countably many such points, so there can only be countably many such triangles.

To put it another way, if there were uncountably many such triangles, then we could pick a different point from the interior of each triangle with rational coordinates, and that would give us uncountably many different points with rational coordinates, which can't happen.

Notice that this argument works for any disjoint collection of sets in $\Bbb R^n$, each of which has a non-empty interior.

Robert Shore
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    What is a "triangle"? The set of the three vertices or it is a "solid triangle", the part of the plane given by intersection three half-planes?! – dan_fulea Sep 06 '19 at 00:30
  • So we can easily use $\mathbb{Q}$'s countablilty and combine in with "every rational number $a$, $b$ has at least on irrational number between them" and visa verse's proof.

    Thanks a lot!

     – Bekir Kırkıcı Sep 06 '19 at 00:31
  • @dan_fulea I assumed (apparently correctly) that it's the interior of the triangle (possibly including the boundary). – Robert Shore Sep 06 '19 at 00:32
  • @BekirKırkıcı Glad I could help. Upvotes and acceptances of answers you find useful are always appreciated. – Robert Shore Sep 06 '19 at 00:33
  • @dan_fulea Sorry I just assumed that it was clear enough that its the interior of a triangle rather than a set of three vertices, thank you for the heads up, I ll be careful next time. – Bekir Kırkıcı Sep 06 '19 at 00:34