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So I think this is the last problem I have and I'm not thinking I'm doing it properly.

Let $a,b$ be real numbers and suppose for all $\varepsilon \gt 0, a \le b+\varepsilon$. Show that $a \le b$.

Assume $b\lt a$. Then $b+\varepsilon \lt a + \varepsilon$ and $$a \le b+\varepsilon \lt a+\varepsilon$$ If I can show that there is a contradiction I can prove it, but I'm having trouble with the contradiction.

Faffi-doo
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    I suppose that http://math.stackexchange.com/questions/679038/a-leq-b-epsilon-for-all-epsilon0-then-a-leq-b-counter-intuitive could help – sas Feb 17 '14 at 23:50
  • http://math.stackexchange.com/questions/622403/proof-by-contradiction-forall-epsilon-in-mathbbr0a-b-epsilon-t – mle Feb 18 '14 at 00:33
  • Just take the limit as $\epsilon \to 0$ – Lemon Feb 18 '14 at 00:35
  • I didn't want to do the limit yet because the text is self contained and limits have not been discussed. The proof should strictly be algebraic properties and order properties of $\mathbb{R}$ – Faffi-doo Feb 19 '14 at 10:42

4 Answers4

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You could prove the contrapositive of this question:

Suppose that $b<a$. Then there exists $\varepsilon>0$ so that $b+\varepsilon<a$.

Ian Coley
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Since you like a proof assuming $b<a$, let's do that. Consider $b+\frac{a-b}2$, halfway between $b$ and $a$.

msh210
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Since $a\le b+\varepsilon$ for all $\varepsilon$. Then either $a\le b$ or $a>b$. Then we will show that the latter leads a contradiction. For sake of contradiction suppose that $a>b$, then $a-b>0$, and since it holds for all $\varepsilon>0$, then in particular it holds for $0<\varepsilon<a-b$. So, $a\le b+\varepsilon<b+(a-b)=a$, i.e., $a<a$. Contradiction. Thus $a\le b$ as desired.

Now try to show that if $|a-L|\le \varepsilon$ for all $\varepsilon>0$, then $a=L$

Jose Antonio
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Assume that $a>b$. Then we have $a>b+\epsilon$ for $\epsilon=\frac {b-1} \epsilon$

Chung. J
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