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Homework Help: How to prove this?

  1. Dec 9, 2004 #1
    so let X(n) be a sequence. Let Y(n) = inf{x(i) : 1 <= i <= n}.

    show y(n) is decreasing.

    ok, first of all, how do I read this problem?

    do I say in words, Y(n) is equal to the infima of x(i), where i is between 1 and n?

    so what does that mean?? no idea where to start.

    ok, so here is how I usually start. I read each sentence, and try to find out what it is by definition, and then try to connect:

    here is attempt:

    -ok, so X(n) is a sequence, so we have X(1)...X(2)...X(i)....to X(n).
    -and Y(n) is the infima(the lower bound) of X(i), where i could be, and including 1 to n.
    -so can I just say Y(n) = inf{X(i)}, where i is arbitrary and between 1 or n.

    ok, so I need to show that y(n) is a decreasing...meaning y(1) < y(j) < y(n) right?

    ok, so how do I do? I mean, I really have no idea on how else to do this besides defintion by definition, so it takes a while...and I still dont know how to connect....

    edit: ok, so here is my attempt at the sollution.

    we need to prove that y(n) is decreasing. This actually means that y(1) < y(2) < y(i) < y(n) right?

    so by definition, if the inf of x(i) = blah, then y = blah right? so basically, Y(n) is the greatest lower bound of X(n).....

    but what does this say about whether Y decrease or increase?

    edit again, man, after 1 hour, I give up...but here is the closest thing I can think of. so if Y(n) is the infimum, that means it is the greatest lower bound. So if we let Y(1) Y(2)......Y(i)....Y(n), that obviously means it is decreasing and is going to the infimum....right?

    im moving on to the next problem, but can someone clear this up for me?
    Last edited: Dec 9, 2004
  2. jcsd
  3. Dec 10, 2004 #2


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    Y need not be decreasing. It is non-increasing, but it need not be decreasing.

    One thing you know is that the infimum will be an element of the set, since it's finite, i.e. Y(n) will only contain n elements, so it's least upper bound will be the least of those n elements. You can show that Y(n) is non-increasing.

    Let Y(n) = inf{X(1), X(2), ..., X(n)} = X(j) (where 1 < j < n)
    Then Y(n + 1) = inf{X(1), ..., X(n), X(n+1)}
    If X(n + 1) > X(j), what will inf{X(1), X(2), .., X(j), ..., X(n), X(n+1)} be, given X(j) = inf{X(1), .., X(n)}? If X(n + 1) < X(j), what will it be? This should be very simple to do. Essentially, all you're proving is that if you have a bunch of numbers, and x is the smallest of those numbers, then if you add another number to the bunch, the smallest number of the new bunch cannot be greater than x.
  4. Dec 10, 2004 #3


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    NO:) Y decreasing means that y1>=y2>=y3>=......yn.

    Decreasing means that it is not increasing, staying unchanged is also allowed.
    And "a less than b" is denoted by a< b.

    To prove the statement that Y is decreasing when yn=inf{X(i)}, you can use the method of induction.

    If n=1, x1=y1. Right?

    If n=2, x2 might be greater, equal or lower than x1. Y2 will be the lower one from both xi-s. If x2 >=x1, y2=y1. If x2<x1, y2=x2<y1. So the statement is true for n=2.

    Now you assume that it is true for n=k, - yk=inf(X(k)}- ,and try to prove the statement for n=k+1.

    Well, yk=inf{x1,...xk}. If x(k+1) is not lower than the lowest among {x1...xk } than yk+1=yk. If x(k+1) < xi for all i<=k than y(k+1) = x(k+1).

    If x(k+1) < xi for all i<=k then inf{X(k+1)}=x(k+1)<inf{X(k)}, that is y(k+1)< yk. Otherwise y(k+1)=yk.
    The statement was true for k=2, and it is true for all the next positive integers, that is, for any n .

  5. Dec 10, 2004 #4


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    "NO:) Y decreasing means that y1>=y2>=y3>=......yn.

    Decreasing means that it is not increasing, staying unchanged is also allowed.
    And "a less than b" is denoted by a< b. "

    While that convention may be used in a specific context, I would expect that it would be stated.

    Normally, "decreasing" means a< b< c, etc. while "non-increasing" means a<= b<= c, etc.

    Some authors, hopefully making it clear that they are using that convention, use
    decreasing to mean a<=b<= c, etc and "strictly decreasing" to a< b< c etc. but I wouldn't consider that standard.
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