Assuming that E is nonempty and bounded subset of $\mathbb{Z}$ then by definition of the completeness axiom E has a finite infium i. since i $\in E$ and $E \subset \mathbb{Z}$. Then i therefore exists and belongs to E. I think I am missing something.
Show that if E is a nonempty subset bounded of $Z$, then inf E exists and belongs to E
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real-analysis
supremum-and-infimum
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1By $Z$, do you mean $\mathbb{Z}$? – 2017-02-01
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0Yes feel free to correct that – 2017-02-01
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0What is this completeness axiom you speak of? you mean that $(\mathbb{Z},d)$ with the usual metric is a complete metric space? – 2017-02-02
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0If E is a nonempty subset of $R$ the real numbers, that is bounded above then E has a finite supremum. I assume this is true for infinum. Also another theorem is if e is contained z then sup E $\in$ E. if the supremeum of a set which only contains integers exists then the at supremum must be an integer – 2017-02-02