# Showing $a \le b$ if $a \le b+\varepsilon$, for all $\varepsilon \gt 0$ [duplicate]

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.

• I suppose that math.stackexchange.com/questions/679038/… could help
– sas
Feb 17, 2014 at 23:50
• math.stackexchange.com/questions/622403/…
– mle
Feb 18, 2014 at 0:33
• Just take the limit as $\epsilon \to 0$ Feb 18, 2014 at 0:35
• 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 $\mathbb{R}$ Feb 19, 2014 at 10:42

Since you like a proof assuming $b<a$, let's do that. Consider $b+\frac{a-b}2$, halfway between $b$ and $a$.
Suppose that $b<a$. Then there exists $\varepsilon>0$ so that $b+\varepsilon<a$.
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 $a=L$
Assume that $a>b$. Then we have $a>b+\epsilon$ for $\epsilon=\frac {b-1} \epsilon$