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Math Help - Compact metric

  1. #1
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    Compact metric

    I know I have to check 3 properties of a metric space, but I don't know how to approach it here. Could somebody help?
    Let M be the set of all bounded sequences ( x_n) of real numbers:
    1) Show that d( ( x_n), ( y_n) )= sup {absolute value of x_n - y_n: n in N } defines a metric on M
    2) Prove that (M,d) is not compact.
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    I know I have to check 3 properties of a metric space, but I don't know how to approach it here. Could somebody help?
    Let M be the set of all bounded sequences ( x_n) of real numbers:
    1) Show that d( ( x_n), ( y_n) )= sup {absolute value of x_n - y_n: n in N } defines a metric on M
    2) Prove that (M,d) is not compact.
    Where in part 1 are you having trouble? That is an exercise in absolute value.

    For part 2, consider the set of all sequences \left\{ {x_n } \right\} in which there are zeros everywhere except a 1 in the nth position.
    Now think about theorems about compactness with regard to that set.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Where in part 1 are you having trouble? That is an exercise in absolute value.

    For part 2, consider the set of all sequences \left\{ {x_n } \right\} in which there are zeros everywhere except a 1 in the nth position.
    Now think about theorems about compactness with regard to that set.
    The way the metric is defined in supremum term threw me off. I didn't know if I can prove this just as it's with absolute value.Here are my arguments:
    a) Since ( x_n) and ( y_n) are bounded and monotonic in R, they both converge.
    d( x_n, y_n)=sup abs ( x_n - y_n) greater than or equal to 0, for all x_n, y_n since absolute value of x_n - y_n is greater than or equal to 0, and the least upper bound must also be greater than or equal to 0.
    Now for the supremum to be equal 0, the absolute value of x_n - y_n must be 0, which means x_n = y_n
    b) Similarly, sup of abs x_n - y_n = sup abs y_n - x_n
    c) I was thinking of using the triangle inequality for this one, but I don't know if it works analogously as it works for absolute value without the supremum.

    For part 2) I'm thinking about the closed and bounded properties of the set you mention, but I didn't get anywhere.
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    For the last question look back at the set of sequences I suggested.
    Here are the first three terms: x_1  = \left( {1,0,0,0, \cdots } \right),\quad x_2  = \left( {0,1,0,0,0 \cdots } \right),\quad x_3  = \left( {0,0,1,0,0, \cdots } \right).
    Is it clear to you that d\left( {x_1 ,x_2 } \right) = 1,\quad d\left( {x_1 ,x_3 } \right) = 1,\quad d\left( {x_3 ,x_2 } \right) = 1?
    Every term in that sequence is one unit from any other term in the set.
    But that is a bounded, infinite set. Can it have a limit point?
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  5. #5
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    Quote Originally Posted by Plato View Post
    For the last question look back at the set of sequences I suggested.
    Here are the first three terms: x_1  = \left( {1,0,0,0, \cdots } \right),\quad x_2  = \left( {0,1,0,0,0 \cdots } \right),\quad x_3  = \left( {0,0,1,0,0, \cdots } \right).
    Is it clear to you that d\left( {x_1 ,x_2 } \right) = 1,\quad d\left( {x_1 ,x_3 } \right) = 1,\quad d\left( {x_3 ,x_2 } \right) = 1?
    Every term in that sequence is one unit from any other term in the set.
    But that is a bounded, infinite set. Can it have a limit point?
    It is clear to me that every term in that sequence is one unit from another term in the set. The set doesn't have a limit point since the term get more distant from the first term infinitely, I think. If it doesn't have a limit point, then it can't be closed.
    Are my arguments in part 1) correct? If not, can you help by pointing out some mistake I made?
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