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seen Sep 29 at 10:28

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....
Aug
24
asked Rudin's PMA Exercise 2.18 - Perfect Sets
Aug
24
comment Convergence of a sequence implies the convergence of another sequence
OP wrote $\sum\limits_{k \text{ is good }} a_k$ which pre-supposes such a sequence exists. If you can't produce an (infinite) sequence then the problem is even easier to solve... In the above "proof" I have left out some details but the gaps are fairly obvious.