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Dick
#6
Jul3-07, 03:38 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
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. 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.