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I am working on a formal proof of the upper bound property of non-empty subsets of N:

Every non-empty subset of N that is bounded from above has within it, the least upper bound of that set.

I can't seem to get anywhere. Here are my thoughts so far, such as they are:

Let x be a non-empty subset of N. Let b be an upper bound of x. Suppose to the contrary that for every element of x, there is a still larger element in x. I should be able to obtain a contradiction from this, but how? Should I consider another approach? Any help would be appreciated.

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    Is N the natural numbers? I may be wrong, but any set of natural numbers which is bounded above is necessarily finite. Could you then maybe use induction to prove that any nonempty finite set of natural numbers has a maximum element?2011-06-06
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    Does $N = \mathbb{N}$? If so, then if $b$ is a least upper bound for a set $S,$ but $b \notin S,$ then $b-1$ is an upper bound for $S.$2011-06-06

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