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 Apr15 awarded Yearling Jul2 awarded Curious Jun30 comment Completeness and Fourier series convergence "Fourier series is about expanding a periodic function." What do you mean by this exactly? Apr15 awarded Yearling Jan10 awarded Popular Question Oct5 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) Oct5 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. Oct5 revised What does it mean to be a “closed subset of a metric space”? edited body Oct5 comment An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Very clear example. Thanks for you answer Andre. Oct5 accepted An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Oct4 revised An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear deleted 25 characters in body Oct4 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. Oct4 revised An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear added 12 characters in body Oct4 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 Oct4 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 Oct4 asked An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear Oct4 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 Oct4 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 Aug24 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.... Aug24 asked Rudin's PMA Exercise 2.18 - Perfect Sets