Ach it'll involve some considerable LaTeX. Just try induction, or even just look at k = 1, 2, 3. If you're still stuck I'll write it out(Original post by ThatPerson)
Though can you explain this step; not sure I quite follow (in particular the second term on the last line):
Ah nice!(Original post by atsruser)
I did this:
Haha, nah just verlan. And yes I'm French.Right. I was wondering if "cimer" was what people did when they got to the top of a mountain or something. Are you bilingual?

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This is cute.
Problem 584:
Is the set of real numbers such that the distance from to the nearest integer goes to 0 as for some positive countable? 
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 01092017 01:06
(Original post by Zacken)
... 
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 27092017 14:25
Problem 585: If is countably infinite, show there exists an uncountable collection of subsets of such that each is infinite and is at most finite for .
Found this in a functional analysis text, though the proof is elementary. As an aside, it can be used to show that there does not exist a bounded projection , which in turn implies that is not isomorphic to a dual space. 
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 27092017 22:52
(Original post by Lord of the Flies)
Ah I remember being fascinated by these numbers
(Original post by ctrls)
Problem 585: If is countably infinite, show there exists an uncountable collection of subsets of such that each is infinite and is at most finite for .
Found this in a functional analysis text, though the proof is elementary. As an aside, it can be used to show that there does not exist a bounded projection , which in turn implies that is not isomorphic to a dual space.
For every pick a sequence of rationals converging to . For two sequences and sharing infinite many terms, these terms would define a common subsequence and so . So we get an uncountable collection with pairwise finite intersections.
Don't know enough functional analysis to understand your last paragraph, but it sounds cool: is set of (bounded?) sequences equipped with the sup norm and is the set of sequences converging to , right? What's a bounded projection and why does the above imply one doesn't exist? What's a dual space? Why does the above imply is not isomorphic to a dual space? Initial thought was that all dual spaces were isomorphic to but that doesn't seem anywhere near correct just via cardinality considerations.Last edited by Zacken; 03102017 at 16:26. 
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 28092017 07:47
(Original post by Zacken)
Biject to the rationals via countability, so it is enough to consider (just invert/transport everything back along the bijection).
For every pick a sequence of rationals converging to . For two sequences and , they would define a common subsequence and so . So we get an uncountable collection with pairwise finite intersections.
(Original post by Zacken)
Don't know enough functional analysis to understand your last paragraph, but it sounds cool: is set of (bounded?) sequences equipped with the sup norm and is the set of sequences converging to , right? What's a bounded projection and why does the above imply one doesn't exist? What's a dual space? Why does the above imply is not isomorphic to a dual space? Initial thought was that all dual spaces were isomorphic to but that doesn't seem anywhere near correct just via cardinality considerations.
Proving this claim takes a bit more work, but the idea is to apply the result to and show that if then the set is countable ( is the indicator function/sequence of each ). If such a projection exists, then , where each is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some , which is a contradiction as each .
The last claim follows from a general fact that if is a normed space, there exists a bounded projection . The idea is to take the canonical embedding and show its adjoint is a projection making appropriate identifications. We can apply this result here since , making sure everything still works up to isomorphism. 
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 28092017 10:18
(Original post by ctrls)
You're correct about the definitions of the spaces (and yes we only consider bounded sequences). The notation is a bit sloppy, but viewing , a projection in this context is a bounded linear map such that and . Also in the context of functional analysis we require all maps to be continuous, so it's much harder for spaces to be isomorphic unlike in linear algebra (we need a linear homeomorphism). In the same spirit, if is a normed space then its dual is the set of continuous linear functionals or .
Proving this claim takes a bit more work, but the idea is to apply the result to and show that if then the set is countable ( is the indicator function/sequence of each ). If such a projection exists, then , where each is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some , which is a contradiction as each .
The last claim follows from a general fact that if is a normed space, there exists a bounded projection . The idea is to take the canonical embedding and show its adjoint is a projection making appropriate identifications. We can apply this result here since , making sure everything still works up to isomorphism. 
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 28092017 12:05
(Original post by Zacken)
Thanks for that; I understood the general gist of it. Functional Analysis seems really cool, definitely something I'll take next year. What book are you using, by the way? 
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 03102017 01:06
Problem 586
To every real assign a finite set of reals that does not contain it. Is there always a continuumsized set that does not intersect its corresponding assigned set of numbers?Last edited by Lord of the Flies; 03102017 at 07:44. Reason: Zacken posted a similar solution to 585 so removed mine. 
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 03102017 03:17
Problem 587*
There are 20 runners in a race. How many different possible podiums are there? 
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 06102017 00:08
These are two problems which have really gorgeous proofs.
Problem 588*
Put finitely many points in the plane, scattered such that they are not all in a line. Show that there is a line going through exactly two of them.
Problem 589*
People attend a party, some are friends some are not. Show that it is possible to divide the people into two groups such that within each group, each person has evenly many friends. 
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 06102017 00:14
(Original post by Lord of the Flies)
Problem 588*
Put finitely many points in the plane, scattered such that they are not all in a line. Show that there is a line going through exactly two of them. 
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 06102017 00:17
(Original post by Zacken)
I've seen the proof of this before, so won't answer it here. But it really is gorgeous (especially considering when you take into account when it was proved). 
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 06102017 00:35
Problem 590***
Given points in the interior of a circle of unit radius, show that there must be a subset of points whose diameter (The maximum distance between any pair of points) is less than .

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 07102017 21:31

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 07102017 22:05
by multiplying by , expanding out a geometric series and interchanging summation and integration using the monotone convergence theorem, integration by parts and quoting the Basel respectively. 
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 07102017 22:28
(Original post by I hate maths)
Problem 591**
Here's one I came across recently. Evaluate .
A hint for A Level students:

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 07102017 22:30
(Original post by _gcx)
Where did you stumble upon this, out of interest? I also found it in a book that I was reading. There's actually a whole thread dedicated to tricky integrals, if you're not already aware 
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 07102017 22:33
(Original post by I hate maths)
Thank you for the link. I got this question from someone I know who's doing maths at university. 
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 20102017 16:17
(Original post by ctrls)
You're correct about the definitions of the spaces (and yes we only consider bounded sequences). The notation is a bit sloppy, but viewing , a projection in this context is a bounded linear map such that and . Also in the context of functional analysis we require all maps to be continuous, so it's much harder for spaces to be isomorphic unlike in linear algebra (we need a linear homeomorphism). In the same spirit, if is a normed space then its dual is the set of continuous linear functionals or .
Proving this claim takes a bit more work, but the idea is to apply the result to and show that if then the set is countable ( is the indicator function/sequence of each ). If such a projection exists, then , where each is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some , which is a contradiction as each .
The last claim follows from a general fact that if is a normed space, there exists a bounded projection . The idea is to take the canonical embedding and show its adjoint is a projection making appropriate identifications. We can apply this result here since , making sure everything still works up to isomorphism.
I haven't figured out the problem you set, alas. I have countablymany pairwisedisjoint subsets of the naturals.
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