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Suppose $ S=\{m\in \mathbb{Z} | y \lt m \}, y \in \mathbb{R}$ . This set is bounded below by $y$. And it is nonempty since Archimedian Property guarrantes the existence of a natural no $n$ which is bigger than $y$. So it has an infimum (Call it $s_*$). I need to show that $s_*$ is an integer i.e I need to show that $s_* \in S$.

If $S$ doesn't contain any non-negative integer then $s_*=0$ if $0 \in S$ or by well ordering property it is a natural no if $0 \not \in S$.

else consider $T=\{m \lt 0| m \in S\}$. There are only finitely many integers which are greater than $s_*$ and less than $0$. I take the minimum of these integers and claim that the minimum element is the infimum

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  • $\begingroup$ Just show that $s_* = \lceil y \rceil$ if $y$ itself in non-integer and $y+1$ otherwise. $\endgroup$ – Stefan Mesken Aug 12 '14 at 5:02
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I don't know why you consider $0$ separately, this seems to be more confusing than anything else.

Your idea with the minimum seems okay, but let me make it look better. Pick $x \in S$. Then $y \le s_* \le x$ and there are finitely many integers between $s_*$ and $x$. Pick $z$ to be the minimum integer between $s_*$ and $x$. Then $z = s_*$, for if $s_* < z$, then $s_* \notin S$ and $z \in S$, so $\frac{s_* + z}2$ is also a lower bound of $S$ (by minimality of $z$) which is greater than $s_*$.

As mentioned in the comments, you should notice that $s_* = y+1$ if $y \in \mathbb Z$ and $s_* = \lceil y \rceil$ otherwise.

Hope that helps,

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Suppose that $s_*\notin S$, then by the axiom (of $\inf$), for $\epsilon =\frac{1}{4}$ there is $s_1\in S$ such that $s_*<s_1<s_*+\frac{1}{4}$, Now there is $s_2\in S$ such that $s_*<s_2<s_1$ (because $s_*<s_1$ take $\epsilon=s_1-s_* $) , it is impossible because $s_1,s_2\in \Bbb Z$ and $|s_2-s_1|\leq \frac{1}{4}$.

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  • $\begingroup$ why is there an $s_2$?? $\endgroup$ – tattwamasi amrutam Aug 12 '14 at 7:32
  • $\begingroup$ $s_1=s_*+\epsilon $ where $\epsilon =s_1-s_*>0$, and axiom of $\inf$. $\endgroup$ – Hamou Aug 12 '14 at 7:34

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