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neutrino
neutrino is offline
#7
Jul3-07, 04:20 PM
P: 2,048
Ah...
Assume inf S > x. This implies that inf S belongs to S.

But then x<(inf S + x)/2 < inf S and it also belongs to S, which would imply inf S is not the infimum. A contradiction!

But trying to prove that x = inf S this way seems redundant since it relies on the fact that for all real x and y, x<(x+y)/2<y, which was what I set to prove in the first place.

And that's what I think you tried to convey in your last post.