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    (Original post by DJMayes)
    Solution 485

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    Let  y \in G . Then by definition  \exists r>0 such that  B(y, r) \subseteq G . Given this, we can easily construct a (possibly uncountable) set of open intervals, the union of which is G:

     \cup_{y \in G} B(y,r) = G

    Now, we take this set and construct a new one by the following method: If two intervals are not disjoint, then their union is an interval; take this union in all possible cases. The result is a set of pairwise disjoint open intervals equal to our set. It remains to prove that this set is countable; to do this we note that our set of intervals injects into the rationals and we are done.

    This questions feels very simple if you know the expected content but due to that I am wary I may have made a mistake here somewhere.

    It feels like you're using the Axiom of Choice here to select your pairwise disjoint open intervals. I'd much rather you didn't :P
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    (Original post by Smaug123)
    It feels like you're using the Axiom of Choice here to select your pairwise disjoint open intervals. I'd much rather you didn't :P
    I'm not sure how to prove this without choice (or at least countable choice). I may not even be true without it.
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    A nice way of doing 485, which doesn't involve the axiom of choice, is setting up an equivalence relation.

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    Exclude the trivial case of G = \emptyset and for g_1,g_2 \in G define the equivalence relation g_1 \sim g_2 if and only if [g_1,g_2] \subseteq G (or the other way around if g_2 < g_1). Then the equivalence classes are precisely pairwise disjoint open intervals in \mathbb{R}. Then from using properties of the equivalence relation, the union of the equivalence classes gives G, but also for each equivalence class we can pick a unique rational (because the intervals must be disjoint from the relation) and so we can find an injective map from the set of equivalence classes into \mathbb{Q} - so the set of equivalence classes is countable, which gives the result.
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    (Original post by Noble.)
    A nice way of doing 485, which doesn't involve the axiom of choice, is setting up an equivalence relation.

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    Exclude the trivial case of G = \emptyset and for g_1,g_2 \in G define the equivalence relation g_1 \sim g_2 if and only if [g_1,g_2] \subseteq G (or the other way around if g_2 < g_1). Then the equivalence classes are precisely pairwise disjoint open intervals in \mathbb{R}. Then from using properties of the equivalence relation, the union of the equivalence classes gives G, but also for each equivalence class we can pick a unique rational (because the intervals must be disjoint from the relation) and so we can find an injective map from the set of equivalence classes into \mathbb{Q} - so the set of equivalence classes is countable, which gives the result.
    You used choice in the bolded bit I think.
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    (Original post by james22)
    You used choice in the bolded bit I think.
    I don't think axiom of choice is needed. The rationals are unique from the equivalence relation and then the rationals can be well-ordered without the axiom of choice.
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    (Original post by Noble.)
    I don't think axiom of choice is needed. The rationals are unique from the equivalence relation and then the rationals can be well-ordered without the axiom of choice.
    It may well be possible to pick the rationals without choice, but just saying they exist and so you can pick them is certainly using choice. You would need to give an explicit contruction of each rational somehow which I don't think can be done.
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    (Original post by Noble.)
    I don't think axiom of choice is needed. The rationals are unique from the equivalence relation and then the rationals can be well-ordered without the axiom of choice.
    You are right, choice is not needed. Although it does simplify the argument slightly.
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    Problem 486**

     \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (9-x)}}{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}} \ dx
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    (Original post by ThatPerson)
    Problem 486**

     \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (9-x)}}{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}} \ dx
    Solution 486

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    Use the substitution  y=6-x :

     \Rightarrow \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (9-x)}}{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}} \ dx = \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (3+y)}}{\sqrt{\ln (9-y)} + \sqrt{\ln (3+y)}} \ dy

     \Rightarrow 2I = \int_2^4 1 \ dx = 2

    Thus I = 1.

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    (Original post by arkanm)
    486, just let ln(9-x)=a*ln(3+x)
    Could you elaborate on this at all? Is this a substitution - if so is  a the variable? What form does this turn the integral into, and how does it help you evaluate it? As it is this doesn't solve anything. On the assumption that you have done this and simply haven't typed it up here I would be interested in seeing it as I don't quite see how anything related to the tidbit you've written could work out nicely.
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    (Original post by DJMayes)
    Solution 486

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    Use the substitution  y=6-x :

     \Rightarrow \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (9-x)}}{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}} \ dx = \displaystyle \int^{4}_{2} \dfrac{\sqrt{\ln (3+y)}}{\sqrt{\ln (9-y)} + \sqrt{\ln (3+y)}} \ dy

     \Rightarrow 2I = \int_2^4 1 \ dx = 2

    Thus I = 1.

    That was Putnam 1987/B1 if you're interested.
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    (Original post by ThatPerson)
    That was Putnam 1987/B1 if you're interested.
    That was a Putnam problem? I'm surprised; I thought they tended to be a fair bit harder. The first couple of questions of the 2014 Putnam paper are quite nice ones if you are interested.
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    (Original post by DJMayes)
    That was a Putnam problem? I'm surprised; I thought they tended to be a fair bit harder. The first couple of questions of the 2014 Putnam paper are quite nice ones if you are interested.
    It does seem to be unusually easy. Most Putnam questions are beyond my reach at the moment but I'll see

    Also, after unsuccessfully trying to compute jjpneed1's integral, I thought I'd search online and found this solution, which makes me feel better about my failed attempt.
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    (Original post by ThatPerson)
    It does seem to be unusually easy. Most Putnam questions are beyond my reach at the moment but I'll see

    Also, after unsuccessfully trying to compute jjpneed1's integral, I thought I'd search online and found this solution, which makes me feel better about my failed attempt.
    That looks like a more complicated version of mine. How could I forget it equals  \pi /2 though? What a nice result
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    (Original post by jjpneed1)
    That looks like a more complicated version of mine. How could I forget it equals  \pi /2 though? What a nice result
    Did you do it without complex analysis?
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    (Original post by ThatPerson)
    Did you do it without complex analysis?
    Split the original integral into a sum of two integrals (with same integrand): one from 0 to 1, other from 1 to infinity. Use x->1/x on the second and recombine the integrals. I think I then used integration by parts. From there you have to evaluate some relatively simpler integrals. It took me a while
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    (Original post by jjpneed1)
    Split the original integral into a sum of two integrals (with same integrand): one from 0 to 1, other from 1 to infinity. Use x->1/x on the second and recombine the integrals. I think I then used integration by parts. From there you have to evaluate some relatively simpler integrals. It took me a while
    Ah, that seems like a nice way of doing it.
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    Problem 487***

    f:\;(0,1)\to (0,1) satisfies f(x)<x. Let \hat{f}(x) =\displaystyle \lim_{n\to\infty}\underbrace{f(f  (\cdots f}_n(x)\cdots); can \hat{f} take uncountably many values?
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    (Original post by Lord of the Flies)
    Problem 487***

    f:\;(0,1)\to (0,1) satisfies f(x)<x. Let \hat{f}(x) =\displaystyle \lim_{n\to\infty}\underbrace{f(f  (\cdots f}_n(x)\cdots); can \hat{f} take uncountably many values?
    This seems too easy. yes since there are uncountably infinite real numbers between 0 and 1. Seems so easy I feel like I've missed something, or at least f(x) can take uncountably many values.
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    (Original post by Jammy Duel)
    This seems too easy. yes since there are uncountably infinite real numbers between 0 and 1. Seems so easy I feel like I've missed something, or at least f(x) can take uncountably many values.
    Can you contruct such an f?

    I've had a go and there are lots of annoying things happening.
 
 
 
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