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Math Help - Bounded Sets

  1. #1
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    Bounded Sets

    Suppose I have two bounded subsets S and T of \mathbb{R} with s \leq t, \forall s \in S and \forall t \in T. Furthermore, suppose the max (S) and min (T) do not exist.

    Is it possible that inf (T) \leq sup (S)?

    Here, I am thinking: Either S \cap T \not = \varnothing or (S \cap T) = \varnothing.
    For (S \cap T) = \varnothing, it's obvious that sup(S) \leq inf(T),
    but for S \cap T \not = \varnothing, it's a little tricky.

    Say I have S=(-20,3) and T=(0,12).
    I find the definition s \leq t, \forall s \in S and \forall t \in T very difficult to understand.
    Since (S \cap T) = [0,3], it looks like s \leq t, \exists s \in S, \exists t \in T
    Last edited by novice; September 9th 2010 at 08:19 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by novice View Post
    Suppose I have two bounded subsets S and T of \mathbb{R} with s \leq t, \forall s \in S and \forall t \in T. Furthermore, suppose the max (S) and min (T) do not exist.

    Is it possible that inf (T) \leq sup (S)?
    Yes, sure. Consider, for example, S := \left\{-\frac{1}{n} \mid n\in\mathbb{N}\right\} and T := \left\{\frac{1}{n}\mid n\in\mathbb{N}\right\}, or maybe even simpler S := \,[-1,0),\; T := \,(0,1]. In that case we have \inf(T)=0=\sup(S).

    But it is impossible for \inf(T) < \sup(S) to hold, I think.
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  3. #3
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    Quote Originally Posted by Failure View Post

    But it is impossible for \inf(T) < \sup(S) to hold, I think.
    Most likely you are right.

    s\leq t, \forall s\in S, \forall t \in T must mean no t \in T is smaller than all the  s \in S, and no s \in S is larger than all the  t \in T, which fit S \cap T= \varnothingeasily.

    Let's say S \cap T \not = \varnothing and s \leq t, \forall s \in S, \forall t \in T. I think this can only be true for two cases:

    Case 1: s \leq t, \forall s \in S, \forall t \in T/S

    Case 2: s \leq t, \forall s \in S/T, \forall t \in T

    If so inf (T) \not < sup (S)

    Then sup (S)  \leq inf (T) holds for S \cap T =\varnothing and S \cap \not = \varnothing

    Yah?
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  4. #4
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    Quote Originally Posted by novice View Post
    Most likely you are right.

    s\leq t, \forall s\in S, \forall t \in T must mean no t \in T is smaller than all the  s \in S, and no s \in S is larger than all the  t \in T, which fit S \cap T= \varnothingeasily.
    No - \forall s \in S \ \forall t \in T \ s \le t means that if you take any  s \in S, then it is smaller (or equal to) than all t \in T, and vice versa.

    Let's say S \cap T \not = \varnothing and s \leq t, \forall s \in S, \forall t \in T. I think this can only be true for two cases:

    Case 1: s \leq t, \forall s \in S, \forall t \in T/S

    Case 2: s \leq t, \forall s \in S/T, \forall t \in T

    If so inf (T) \not < sup (S)

    Then sup (S)  \leq inf (T) holds for S \cap T =\varnothing and S \cap \not = \varnothing

    Yah?
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    No - \forall s \in S \ \forall t \in T \ s \le t means that if you take any  s \in S, then it is smaller (or equal to) than all t \in T, and vice versa.
    Any s \in S can easily be smaller than all the t \in T, but for any s \in S to be equal to all the t \in T, then |S|=|T|=1.
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