Okay, so I think I kind of get this one already. Since 2 is the lowest rational number in the set that's less than $x$, then $\inf S = 2$.
But is there is any other way to explain this? I feel like there's a more formal proof to it that I'm not getting. Or maybe I'm just completely wrong. Can someone help out here?
On
In order to prove that $S = 2$, we must show that $S > 2$ and $S < 2$ cannot be infima of this set.
Consider $S < 2$. Now there exists some $x \in \mathbb{Q}$ such that $S < x < 2$, and hence we have found a greater lower bound than our $S$, this is a contradiction hence $S \not < 2$.
Consider $S > 2$, can you take it from here?
By definition of $ S $, $2$ is a lower bound for $ S $. On the other hand, for every $\epsilon > 0$, $2+\epsilon $ is not a lower bound for $ S $. Indeed, given $\epsilon > 0$, there exists an $ n\in \Bbb N$ such that $\frac{1}{n} < \epsilon$. So $2 + \frac{1}{n} $, an element of $ S $, is less than $2+\epsilon $. Therefore, $\inf S = 2$.