In a real analysis problem, I assumed directly that for all the integers greater than a real number, there is an integer closest to that number. I want to somehow justify this. Is this a generalization of the well-ordering principle? Is it an axiom?
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1It is helpful to approach questions like this by first taking a close look at the definitions of key terms like "minimum" and "bounded below". – anak Jun 18 '15 at 13:06
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Let $r$ be your real number and $I$ the set of integers, i.e. $$ I = \{ n\in \mathbb{Z} | n\geq r\}. $$ It is clear that $I$ is nonempty. Suppose that $I$ does not have a minimum: this means that for all $m\in I $ there exists an $p\in I$ such that $ m>p$. This means that if $m\in I$, also $m-1 \in I$. Since $r$ is finite, there exists an integer $l$ such that $l<r$. Let $m\in I$, then by applying the argument above $m-l$ times we see that also $l\in I$. This is a contradiction, implying that $I$ must have minimum.
Hrodelbert
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I'll try to give a proof of the fact that every nonempty set of integers that is bounded below has a smallest element. Hope this is reasonable.
Suppose $S$ is an arbitrary nonempty set of integers that is bounded below. First we will show that $\exists m \in S \forall n \in S (m \leq n)$. By the Axiom of Completeness it is easy to check that $S$ has a greatest lower bound. Let $s = \text{inf} \: S$, and $L = \left\{ x\in \mathbb{R} \mid \forall n \in S \left( x \leq n \right) \right\}$ be the set of all lower bounds for $S$. Clearly $s \in L$. It is not difficult to show that $s + 1 \notin L$. Then we can find some $m \in S$ such that $s + 1 > m$, so $s \leq m < s + 1$. Now suppose $n \in S$ and assume for the sake of contradiction that $m > n$. Then $s + 1 > m > n$ and $n \geq s$ imply that $s \leq n < s + 1$. Hence, since $m,n \in \mathbb{Z}$, we have that $m = n$, which is a contradiction. Thus our assumption that $m > n$ is false, so we must have $m \leq n$. Therefore, since $n \in S$ was arbitrary we get that $\forall n \in S (m \leq n)$, and we can conclude that that choice of $m \in S$ is the smallest element of $S$.
Now we will show that $m = s$. Since $s = \text{inf} \: S$ we get that $s \leq m$. But because $m \in L$ we also have that $s \geq m$. Therefore, $m = s$, as claimed.
brandao
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1$s+1>n$ doesn't imply $s≥n$, for $s∈\mathbb{R}$ and $n∈\mathbb{Z}$; you also need $s∈\mathbb{Z}$, but this was never proven. To argue that $\inf S \in S$, you can modify this proof: supremum of a set of integers is in the set (which I think implicitly used Archimedean Principle). – Yibo Yang Jul 26 '18 at 16:37
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@YiboYang: The proof I gave was badly flawed. Thanks for pointing out. I hope this modified one, though lengthy, is okay. – brandao Jul 31 '18 at 21:02
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By the archimedean property, there exists an integer smaller than that real number, and we are back to the case in $\mathbb{Z}$.
EDIT:
Elaborating:
Let $E$ be the set of integers such that $x \in E \geq r$, $r$ real number. If $E$ is bounded below by an integer, then the case is of a subset of $\mathbb{Z}$ bounded below, which we already know how to deal with.
If $r\geq 0$, $-1$ is such a bound.
If $r< 0 $, $-r>0$ and the archimidean property (applied to $1$ and $-r$) implies that there exists a natural such that $n >-r \implies -n < r$
Aloizio Macedo
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