# Subset Condititonal Question

1. Oct 25, 2010

### Zarlucicil

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.

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

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. Oct 25, 2010

### HallsofIvy

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.

3. Oct 25, 2010

### Zarlucicil

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.

4. Oct 26, 2010

### zut837

This is true for any ordered set S. Just pick s'=s''=t