$x,y \in \mathbb R$ are such that $x \lt y + \epsilon$ for any $\epsilon \gt 0$ Then prove $x \le y$.
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2$\begingroup$ You mean $x<y+\epsilon$ for every $\epsilon>0$, right? Assume the opposite to reach a contradiction. $\endgroup$– SayanNov 18, 2014 at 7:40
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$\begingroup$ for any $\epsilon \gt 0$ $\endgroup$– Highlights FactoryNov 18, 2014 at 7:53
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3$\begingroup$ This isn't a problem of proving something we are uncertain of. This is a problem of using a set of axioms to prove something that had damn well better be true. So to approach this problem, you must specify what your axioms are. Find a reference for an axioma of real numbers. If this problem came from a book, check which theorems the chapter introduced and see if you can use those. $\endgroup$– DanielVNov 18, 2014 at 8:26
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$\begingroup$ What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. $\endgroup$– user1729Nov 18, 2014 at 10:03
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$\begingroup$ See also math.stackexchange.com/questions/679038/… $\endgroup$– Martin SleziakAug 29, 2016 at 4:52
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2 Answers
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If $x > y $, then $ x - y > 0 $. Hence, we can choose $\epsilon = x - y$ and we obtain
$$ x < y + \epsilon = y + x - y = x $$
Contradiction.
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Hints:
If not. Pick $\epsilon=x-y$.
