# Proof that given the info below, sup S <= inf T

1. Jan 30, 2010

### thedoctor818

1. The problem statement, all variables and given/known data
Let S and T be nonempty subsets of $$\mathbb{R} \backepsilon s \leq t \forall s \in S \wedge t \in T.$$ A) Observe that S is bounded above and that T is bounded below. B) Prove that $$sup S = inf T.$$

2. Relevant equations

3. The attempt at a solution
Let $$s_0 = sup S. \text{ Then } s \leq s_o \forall s \in S \wedge s \leq t \forall s \in S \Rightarrow s_0 \leq t.$$
Let $$t_0 = inf S. \text{ Then } t_0 \leq t \forall t \in T \wedge s \leq t \forall t \in T \Rightarrow s \leq t_0.$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 30, 2010

### sutupidmath

I am probbably missing something, but it doesn't seem that it is even true. COnsider
S=(0,1) and T=(2,3). Then for all s in S s=<t for all t in T, but clearly supS=/=infT. Are you sure the problem is not asking you to show : supS=<infT??? If this is what the problem is asking, then try a proof by contradiction. Namely, assume that infT<supS, and derive a contradiction.

Last edited: Jan 30, 2010
3. Jan 31, 2010

### HallsofIvy

Yes. The title says "sup S$\le$ inf T" but in your post you say "sup S< sup T" which does not follow from your hypotheses. They might be equal. If sup S> sup T, then let $\delta$= Sup S- sup T. There must be some member of S closer to S than that and so must be some member of S larger than sup T.