Showing $a \le b$ if $a \le b+\varepsilon$, for all $\varepsilon \gt 0$ 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.
You could prove the contrapositive of this question:


  Suppose that $b&lt;a$. Then there exists $\varepsilon&gt;0$ so that $b+\varepsilon&lt;a$.
Assume that $a&gt;b$. Then we have $a&gt;b+\epsilon$ for $\epsilon=\frac {b-1} \epsilon$