Simple Analysis Question On Upper/Lower Bound of Real Number Let $a$ be real number with $a \geq \varepsilon$ for some $ \varepsilon >0$. Am I allowed to conclude that $a>0$? 
My thinking is "yes" and my reasoning is as follows: Suppose not, then $ a \leq 0$. Now picking $\varepsilon >0$ such that $a\geq \varepsilon$, I obtain
$$
0< a \leq 0
$$
which is absurd. Is my logic above correct or I simply made some mistake in my reasoning.
 A: It's impossible to have $a\geq \varepsilon$ for every $\varepsilon>0$. Indeed, take $\varepsilon:=\max\{1,a+1\}$. Then $\varepsilon>0$ and $\varepsilon\geq a+1>a$.
I think you mean to ask if we have $a\geq \varepsilon$ for some $\varepsilon>0$, then can we conclude $a>0$? The answer to this question is "yes" by the transitivity of "$<$":
$$
0 < \varepsilon \leq a.
$$
A: Proof by contradiction. Assume $a\leq0$ then by Trichotomy Property of R, exactly one is true $a=0$ or $-a>0$.
(i) $a=0$ then $a-\varepsilon=-\varepsilon<0$ which contradicts to the given information that $a-\varepsilon\geq0$
(ii) $-a>0$ then by Order Property of R $-a+\varepsilon>0$ and then $\varepsilon>a$ which again contradicts to the given information.
Thus by contradiction $a>0$
A: "Let a be real number with a≥ε for some ε>0. Am I allowed to conclude that a>0?"
Um... Transitivity.  
You have $a \ge \epsilon > 0$.  So $a > 0$. 
If you really need a proof, below are the neausiating details:
$a \ge \epsilon$ means either $a > \epsilon$ or $a = \epsilon$.
Case 1:  $a > \epsilon$ and $\epsilon > 0$ so $a > 0$.
Case 2: $a = \epsilon$ and $\epsilon > 0$.  So substitute $\epsilon$ for $a$.  $a > 0$.
