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 Apr 26 awarded Notable Question Apr 15 awarded Yearling Jul 2 awarded Curious Jun 30 comment Completeness and Fourier series convergence "Fourier series is about expanding a periodic function." What do you mean by this exactly? Apr 15 awarded Yearling Jan 10 awarded Popular Question Oct 5 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Then there is a "minimal criminal" a_n in S such that a_n is the member of a triple which lies in a straight line and the other two points in the triple are before it in the sequence. But this contradicts the fact that collinearity works for the finite case (Up to a_n) Oct 5 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Andre, on second passing, it looks to me like you've only proven that (a) the collinearity condition holds for every finite case. I am not convinced that you have proven it holds for the infinite case. However, let me try to rectify this. You have proven (a) and also (b) the construction gives us an infinite sequence S whose set of limit points is [0,1]. It remains to be shown that the collinearity condition holds for S. Suppose it doesn't. Oct 5 revised What does it mean to be a “closed subset of a metric space”? edited body Oct 5 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Very clear example. Thanks for you answer Andre. Oct 5 accepted An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Oct 4 revised An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear deleted 25 characters in body Oct 4 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Yes you are right and my OP was wrong before and I just corrected it thanks to your comments. Oct 4 revised An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear added 12 characters in body Oct 4 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Well actually what I meant to write was that [0,1] is a subset of the closure of S but it amounts to the same thing Oct 4 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear njguliyev - why must S be a subset of [0,1]? And Ross yes you are right about the second question of course Oct 4 asked An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Oct 4 accepted $S\subset \mathbb{R}^2$ with one and only one limit point in $\mathbb{R}^2$ such that no three points in $S$ are collinear Oct 4 asked $S\subset \mathbb{R}^2$ with one and only one limit point in $\mathbb{R}^2$ such that no three points in $S$ are collinear Aug 24 comment Rudin's PMA Exercise 2.18 - Perfect Sets Some of what you said makes sense (I understand some of your ideas). Some is unclear to me. And you haven't proven anything concretely. And there are other answers....