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Subset Condititonal Question

  1. Oct 25, 2010 #1
    I've used the following implication (conditional...whatever you want to call it) in a few proofs and was wondering if it's actually is true. I incorporated it into my proofs because it seemed to make obvious sense, but I'm not sure if I'm overlooking something- obvious or subtle.

    [tex] T \subseteq S \Rightarrow \exists s' \in S \& \exists s'' \in S \ni [s' \leq t \leq s''], \forall t \in T [/tex].

    English: If T is a subset of S, then there exists an s' in S and an s'' in S such that t is greater than or equal to s' and less than or equal to s'', for all t in T.
     
    Last edited: Oct 25, 2010
  2. jcsd
  3. Oct 25, 2010 #2

    HallsofIvy

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    First, we are not talking about general sets. In order for the inequalities to make sense, S must be a linearly ordered set- probably the set or real numbers. And it looks to me like, in order for that statement to be true, S and T must be intervals specifically.
     
  4. Oct 25, 2010 #3
    Yes- I'm sorry. S is a subset of the real numbers.
    I suppose that might be true, but I can't think of a counterexample involving non-interval sets nor have I found a way to disprove the implication for non-interval sets. It seems to be true for at least some non-interval sets. For example, when T = {-3.2, -1, 7} and S = {-4, -3.2, -1, 0, 7, 9}. Hmm, or are these example sets considered to be "intervals" because they can be written as the union of intervals? --> T = [-3.2, -3.2] U [-1, -1] U [7, 7]. If they are considered to be intervals, then I don't know what wouldn't be considered an interval.
     
  5. Oct 26, 2010 #4
    This is true for any ordered set S. Just pick s'=s''=t
     
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