Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    18
    ReputationRep:
    (Original post by ThatPerson)
    Though can you explain this step; not sure I quite follow (in particular the second term on the last line):
    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 atsruser)
    I did this:
    Ah nice!

    Right. I was wondering if "cimer" was what people did when they got to the top of a mountain or something. Are you bilingual?
    Haha, nah just verlan. And yes I'm French.

    :smokin:
    Offline

    22
    ReputationRep:
    This is cute.

    Problem 584:

    Is the set of real numbers \alpha > 1 such that the distance from k\alpha^n to the nearest integer goes to 0 as n\to \infty for some positive k countable?
    Offline

    18
    ReputationRep:
    (Original post by Zacken)
    ...
    Ah I remember being fascinated by these numbers
    Offline

    9
    ReputationRep:
    Problem 585: If S is countably infinite, show there exists an uncountable collection \{A_i\}_{i \in I} of subsets of S such that each A_i is infinite and A_i \cap A_j is at most finite for i \neq j.

    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 P : \ell_{\infty} \rightarrow c_0, which in turn implies that c_0 is not isomorphic to a dual space.
    Offline

    22
    ReputationRep:
    (Original post by Lord of the Flies)
    Ah I remember being fascinated by these numbers
    Sorry, didn't see this reply earlier for some reason. They do have some pretty properties, but I don't know much about them.

    (Original post by ctrls)
    Problem 585: If S is countably infinite, show there exists an uncountable collection \{A_i\}_{i \in I} of subsets of S such that each A_i is infinite and A_i \cap A_j is at most finite for i \neq j.

    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 P : \ell_{\infty} \rightarrow c_0, which in turn implies that c_0 is not isomorphic to a dual space.
    Biject to the rationals via countability, so it is enough to consider \mathbb{Q} (just invert/transport everything back along the bijection).

    For every \alpha \in \mathbb{R} pick a sequence of rationals A_{\alpha} = \{\alpha_n \}_{n\in \mathbb{N}} converging to \alpha. For two sequences \alpha_n and \beta_n sharing infinite many terms, these terms would define a common subsequence and so \alpha = \beta. So we get an uncountable collection \{A_{\alpha}\}_{\alpha \in \mathbb{R}} with pairwise finite intersections.

    Don't know enough functional analysis to understand your last paragraph, but it sounds cool: \ell_{\infty} is set of (bounded?) sequences equipped with the sup norm and c_0 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 c_0 is not isomorphic to a dual space? Initial thought was that all dual spaces were isomorphic to \ell_{0} but that doesn't seem anywhere near correct just via cardinality considerations.
    Offline

    9
    ReputationRep:
    (Original post by Zacken)
    Biject to the rationals via countability, so it is enough to consider \mathbb{Q} (just invert/transport everything back along the bijection).

    For every \alpha \in \mathbb{R} pick a sequence of rationals A_{\alpha} = \{\alpha_n \}_{n\in \mathbb{N}} converging to \alpha. For two sequences \alpha_n and \beta_n, they would define a common subsequence and so \alpha = \beta. So we get an uncountable collection \{A_{\alpha}\}_{\alpha \in \mathbb{R}} with pairwise finite intersections.
    Perfect, this was also what was given in the book.

    (Original post by Zacken)
    Don't know enough functional analysis to understand your last paragraph, but it sounds cool: \ell_{\infty} is set of (bounded?) sequences equipped with the sup norm and c_0 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 c_0 is not isomorphic to a dual space? Initial thought was that all dual spaces were isomorphic to \ell_{0} but that doesn't seem anywhere near correct just via cardinality considerations.
    You're correct about the definitions of the spaces (and yes we only consider bounded sequences). The notation is a bit sloppy, but viewing c_0 \subset \ell_{\infty}, a projection in this context is a bounded linear map P : \ell_{\infty} \rightarrow \ell_{\infty} such that P^2=P and P(\ell_{\infty}) = c_0. 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 X is a normed space then its dual X^* is the set of continuous linear functionals P : X \rightarrow \mathbb R or \mathbb C.

    Proving this claim takes a bit more work, but the idea is to apply the result to S = \mathbb N and show that if f \in (\ell_{\infty})^{\ast} then the set \{i \in I : f(\chi_{A_i}) \neq 0 \} is countable (\chi_{A_i} is the indicator function/sequence of each A_i). If such a projection P exists, then c_0 = \mathrm{Ker}\, (\mathrm{id}-P) = \bigcap_{n \in \mathbb N} e_n^{\ast} \circ (\mathrm{id}-P), where each e_n^{\ast} \in (\ell_{\infty})^{\ast} is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some \chi_{A_i}, which is a contradiction as each \chi_{A_i} \not\in c_0.

    The last claim follows from a general fact that if X is a normed space, there exists a bounded projection P : X^{***} \rightarrow X^*. The idea is to take the canonical embedding i : X \rightarrow X^{**} and show its adjoint is a projection making appropriate identifications. We can apply this result here since c_0^{**} \cong \ell_{\infty}, making sure everything still works up to isomorphism.
    Offline

    22
    ReputationRep:
    (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 c_0 \subset \ell_{\infty}, a projection in this context is a bounded linear map P : \ell_{\infty} \rightarrow \ell_{\infty} such that P^2=P and P(\ell_{\infty}) = c_0. 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 X is a normed space then its dual X^* is the set of continuous linear functionals P : X \rightarrow \mathbb R or \mathbb C.

    Proving this claim takes a bit more work, but the idea is to apply the result to S = \mathbb N and show that if f \in (\ell_{\infty})^{\ast} then the set \{i \in I : f(\chi_{A_i}) \neq 0 \} is countable (\chi_{A_i} is the indicator function/sequence of each A_i). If such a projection P exists, then c_0 = \mathrm{Ker}\, (\mathrm{id}-P) = \bigcap_{n \in \mathbb N} e_n^{\ast} \circ (\mathrm{id}-P), where each e_n^{\ast} \in (\ell_{\infty})^{\ast} is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some \chi_{A_i}, which is a contradiction as each \chi_{A_i} \not\in c_0.

    The last claim follows from a general fact that if X is a normed space, there exists a bounded projection P : X^{***} \rightarrow X^*. The idea is to take the canonical embedding i : X \rightarrow X^{**} and show its adjoint is a projection making appropriate identifications. We can apply this result here since c_0^{**} \cong \ell_{\infty}, making sure everything still works up to isomorphism.
    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?
    Offline

    9
    ReputationRep:
    (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?
    The book is Carothers' "A Short Course on Banach Space Theory." It assumes prior knowledge of the subject, but it's quite a nice and gentle introduction to some of the later topics.
    Offline

    18
    ReputationRep:
    Problem 586

    To every real assign a finite set of reals that does not contain it. Is there always a continuum-sized set that does not intersect its corresponding assigned set of numbers?
    Offline

    3
    ReputationRep:
    Problem 587*

    There are 20 runners in a race. How many different possible podiums are there?
    Offline

    18
    ReputationRep:
    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.
    Offline

    22
    ReputationRep:
    (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.
    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).
    Offline

    18
    ReputationRep:
    (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).
    Indeed! And an incredibly frustrating problem too. It was an a Numbers & Sets sheet funnily enough — no rewards for guessing who the lecturer was...
    Offline

    11
    ReputationRep:
    Problem 590***
    Given 541 points in the interior of a circle of unit radius, show that there must be a subset of 10 points whose diameter (The maximum distance between any pair of points) is less than \frac{\sqrt{2}}{4}.

    Spoiler:
    Show

    Hint : Superimpose a \frac{1}{4} unit by \frac{1}{4} unit grid over the circle with one of the grid points falling on the center of the circle you should be able to show that the circle is covered by 60 grid squares, then apply the pigeon hole principle
    Offline

    10
    ReputationRep:
    Problem 591**
    Here's one I came across recently. Evaluate \displaystyle \int^\infty_0\frac{x}{e^x-1}\ dx.

    A hint for A Level students:

    Spoiler:
    Show








    \displaystyle \sum^\infty_{i=1}\frac{1}{i^2} = \frac{\pi^2}{6}







    Offline

    22
    ReputationRep:
    (Original post by I hate maths)
    Problem 591**
    Here's one I came across recently. Evaluate \displaystyle \int^\infty_0\frac{x}{e^x-1}\ dx
    \displaystyle

\begin{align*}\int_0^{\infty} \frac{x}{e^x - 1} \, \mathrm{d}x &= \int_0^{\infty} \frac{xe^{-x}}{1 - e^{-x}} \, \mathrm{d}x \\  & = \sum_{n \geqslant 1} \int_0^{\infty}xe^{-nx} \, \mathrm{d}x \\ & = \sum_{n \geqslant 1} \frac{1}{n} \int_0^{\infty} e^{-nx} \, \mathrm{d}x \\ & = \sum_{n \geqslant 1} \frac{1}{n^2} = \frac{\pi^2}{6}\end{align*}

    by multiplying by e^{-x}/e^{-x}, expanding out a geometric series and interchanging summation and integration using the monotone convergence theorem, integration by parts and quoting the Basel respectively.
    Offline

    18
    ReputationRep:
    (Original post by I hate maths)
    Problem 591**
    Here's one I came across recently. Evaluate \displaystyle \int^\infty_0\frac{x}{e^x-1}\ dx.

    A hint for A Level students:

    Spoiler:
    Show










    \displaystyle \sum^\infty_{i=1}\frac{1}{i^2} = \frac{\pi^2}{6}









    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, I'm sure you'll find it interesting
    Offline

    10
    ReputationRep:
    (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
    Thank you for the link. I got this question from someone I know who's doing maths at university.
    Offline

    18
    ReputationRep:
    (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.
    No problem If you're interested, I stumbled across it in a book called "Inside Interesting Integrals", which might be worth checking out if you have an interest in hard integrals. (also Alyafeai's Advanced Integration Techniques is pretty nice)
    Offline

    7
    ReputationRep:
    (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 c_0 \subset \ell_{\infty}, a projection in this context is a bounded linear map P : \ell_{\infty} \rightarrow \ell_{\infty} such that P^2=P and P(\ell_{\infty}) = c_0. 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 X is a normed space then its dual X^* is the set of continuous linear functionals P : X \rightarrow \mathbb R or \mathbb C.

    Proving this claim takes a bit more work, but the idea is to apply the result to S = \mathbb N and show that if f \in (\ell_{\infty})^{\ast} then the set \{i \in I : f(\chi_{A_i}) \neq 0 \} is countable (\chi_{A_i} is the indicator function/sequence of each A_i). If such a projection P exists, then c_0 = \mathrm{Ker}\, (\mathrm{id}-P) = \bigcap_{n \in \mathbb N} e_n^{\ast} \circ (\mathrm{id}-P), where each e_n^{\ast} \in (\ell_{\infty})^{\ast} is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some \chi_{A_i}, which is a contradiction as each \chi_{A_i} \not\in c_0.

    The last claim follows from a general fact that if X is a normed space, there exists a bounded projection P : X^{***} \rightarrow X^*. The idea is to take the canonical embedding i : X \rightarrow X^{**} and show its adjoint is a projection making appropriate identifications. We can apply this result here since c_0^{**} \cong \ell_{\infty}, making sure everything still works up to isomorphism.
    There was a time you would drop these in my Facebook inbox every now and again . Very nice. It blew my mind cos I realised I'd never bothered to understand projections properly. I was thinking "can one project onto any subspace? Sure, why not."

    I haven't figured out the problem you set, alas. I have countably-many pairwise-disjoint subsets of the naturals.
    Spoiler:
    Show

    \{a2^n:n\in\mathbb{N}_0\}, for odd a
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.