Is my proof here correct (basic real analysis/order theory question)?

Question

I'm self-studying real analysis and came across a simple problem I was trying to solve (although I think this is more of an order theory problem):

Let $A\subseteq \mathbb{R}$ so that $ε\:>0$ and $\forall \alpha \in A:\alpha >\:ε$

Show that the greatest lower bound (infimum) of $A$ is not $0$.

Pardon me if the language is somewhat imprecise, I translated this question to English and I'm not yet fully proficient in English mathematical terminology.

Anyway, I solved this question in the following way: since all elements of set $A$ are bigger than $ε$, I figured out that $Inf A=ε$, and since $ε>0$, we can use transitivity to conclude that $Inf A > 0$, and therefore $Inf A \neq 0$.∎

Is my proof correct and acceptable? Because in the solutions file, the author of the question did something different and proved it by contradiction.

Thank you very much :)

Answer 1

As Dave points in a comment, you can conclude that $\inf A \geq \varepsilon$, yielding $\inf A \geq \varepsilon > 0,$ whence $\inf A > 0$.

Why can't you conclude that $\inf A = \varepsilon$?
Well, suppose, for example that $A=\{2\}$ and $\varepsilon=1$.
Then $\inf A = 2$.
Of course you still have that $\inf A > 0$, which is what you wanted to prove.

Answer 2

Your proof is incorrect because you assume $\inf{A} = \epsilon$.

Here's a proof. Suppose $\inf{A} = 0$. Let $\epsilon > 0$. Since $\inf{A}$ is the infimum, there exists an element $\alpha \in A$ such that $\inf{A} + \epsilon = \epsilon > \alpha$, contradicting the hypothesis.