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 P: 2,046 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