This is a question about "taking the inf" that is done everywhere in analysis. I give one example here to make my question more concrete.
In the proof that the Lebesgue outer measure of intervals is equal to the length of the interval $[a,b]$, we take any cover of intervals $\{I_n\}$ and show that
$$\sum_{n=1}^\infty |I_n| \geq b-a,$$
as in equations 1-8 here.
Then, we "take the infinum of this set of numbers" and obtain that
$$m^*([a,b]) \geq b-a.$$
Why is it that we can take the infinum and the inequality still holds? Can we always do this? In the first inequality, we had that for any cover the inequality holds. But the infinum is not referring to a specific cover, it is rather the largest lower bound of all these numbers. Why is it not possible that this largest lower bound is less than $b-a$?
How can this be made more explicit? Perhaps by stating that for every $\epsilon >0$, there exists a specific cover $\{I'_n\}$ such that
$$m^*([a,b]) + \epsilon \geq \sum_{n=1}^\infty |I_n'| $$
and hence because the first inequality holds for any cover and we can take $\epsilon$ to be arbitrarily small,
$$m^*([a,b]) \geq b-a.$$
Is that "how infimums work" in this context?
Your question is this: if we have a set $A$ and a number $\alpha$ such that $(\forall a\in A):a\geqslant\alpha$, then $\inf A\geqslant\alpha$. This is true, because we are assuming that $\alpha$ is a lower bound of $A$ and, by definition, $\inf A$ is the greatest lower bound of $A$.