Real Analysis Convergance

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If E is a subset of R, nonempty, and bounded, then there exists e in E such that supE-1 < e < supE.

I understand how this works for the subset with natural numbers because for the less than or equal to inequality, supE-1= e. However, I do not understand how this can work for all subsets of R where the inequality is strictly less than. I was thinking that this possibly had something to do with the Archimedean principle that for every x in R there is an n in N such that n>x. So there must be a real number between supE-1 and supE.

Any help is appreciated in advance!