 Quote by Dick
Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S.
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?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.
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Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.
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In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.