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$x,y \in \mathbb R$ are such that $x \lt y + \epsilon$ for any $\epsilon \gt 0$ Then prove $x \le y$.

Martin Sleziak
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Highlights Factory
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    You mean $x<y+\epsilon$ for every $\epsilon>0$, right? Assume the opposite to reach a contradiction. – Sayan Nov 18 '14 at 07:40
  • for any $\epsilon \gt 0$ – Highlights Factory Nov 18 '14 at 07:53
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    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. – DanielV Nov 18 '14 at 08:26
  • 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. – user1729 Nov 18 '14 at 10:03
  • See also http://math.stackexchange.com/questions/679038/intuition-if-a-leq-b-epsilon-for-all-epsilon0-then-a-leq-b – Martin Sleziak Aug 29 '16 at 04:52

2 Answers2

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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 $\epsilon=x-y$.

Paul
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