I'll address your two main concerns in order. First, "Why do we need to show that 0 is the greatest lower bound?"

At the point where the proof stops making sense to you, we've shown that every element of the set E is greater than 0. Thus, it is

*a* lower bound. But can we be sure there isn't some other number, greater than 0, such that every element of E is also greater than that number? Suppose for a moment the sequence was [tex]F = \left\{\dfrac{1}{2n} \bigg \rvert n \in \mathbb{N}^* \right\}[/tex]. As before, it's definitely true that every element of F is greater than 0, so 0 is a lower bound. However, n must be a strictly positive

real number. If we plug in n = 1, we can see that the smallest possible element in F is 1/2. This means that 0

*isn't* the infimum of F, even though it is a lower bound.

Second, "What is the proof doing? I don't understand it."

To show that 0 is the greatest lower bound, and thus the infimum, we need to show that, if we pick

*any positive number* (this is the [tex]\epsilon[/tex]), there is an element of E that's bigger than 0 but less than our choice of [tex]\epsilon[/tex]. What your professor did in the proof was to split up the sequence defining E into two separate sequences. We know that [tex]\dfrac{1}{n}[/tex] races off towards 0 as n increases towards infinity, and the same applies for [tex]\dfrac{1}{m}[/tex]. Thus, by picking a big enough value for n, we can guarantee that [tex]\dfrac{1}{n} < \epsilon[/tex], no matter how small an [tex]\epsilon[/tex] we pick. In fact, since n can be as big as we need, [tex]\dfrac{1}{n}[/tex] can be made even smaller than [tex]\dfrac{\epsilon}{2}[/tex].

Say we picked [tex]\epsilon = 10^{-33} = 0.000000000000000000000000000000001[/tex]. That's super duper crazy small! But, even so, we can still pick a big enough n to satisfy the criteria. Specifically, we can pick, say, [tex]n = 10^{100}[/tex], because [tex]\dfrac{1}{n} = \dfrac{1}{10^{100}} = 10^{-100} \ll \dfrac{10^{-33}}{2}[/tex]

Then, because both [tex]\dfrac{1}{n}[/tex] and [tex]\dfrac{1}{m}[/tex] can be made smaller than [tex]\dfrac{\epsilon}{2}[/tex], we can definitely guarantee that their sum is smaller than [tex]\epsilon[/tex]. And thus, for

*any* choice [tex]\epsilon > 0[/tex], there exists an element of E smaller than it, and so we've proven there cannot exist a lower bound of E except 0. We therefore know that 0 is the infimum of E.

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