Re: analysis exercise,



On Wed, 30 Aug 2006, vsgdp wrote:

Let S and T be nonempty subsets of R with the property that s <= t for all s
in S and all t in T. Prove that sup S <= inf T.

I see this is true from a picture but I could not come up with a direct
proof. I tried supposing sup S > inf T. Then pictorially, there should be
an s in S greater than a t in T to give the contradiction. But I am having
a hard time translating the picture into math symbols.

T is nonempty, so pick any t1 in T. Is it an upper bound for the set S?
Now sup S is the least upper bound for S, so sup S <= t1.

Then sup S is a lower bound for T, so ... (you can now finish it).

Cheers, ZVK(Slavek).
.

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