What is the reasoning behind this rigorous proof of the infimum of $A$?

Question

Take the set

$$A = \left\{\frac{n^2-1}{n^3-1} : n \in \mathbb{N}, n \neq 1 \right\}.$$

I'm proving that $\inf A = 0$. I've already shown the lower bound is $0$, so the only thing left to do is show any other lower bound of $A$ is less than or equal to $0$.

If $A$ was dense, I would pick the middle point between $0$ and some lower bound $z$ bigger than $0$ and show there exists an element smaller than $z$ but this isn't the case so I'm stuck

The solution shows a rigorous method I haven't seen before:

• We prove for all $\varepsilon > 0$ there exists $n \in \mathbb{N}, (n \neq 1)$ such that $\frac{n^2-1}{n^3-1} = \frac{n+1}{n^2 + n + 1} < \varepsilon$. The limit was calculated to be $0$ as $n$ tends to infinity, which is clearly smaller than any positive $\varepsilon$

I'm having trouble understanding intuitively how this proves that $0$ is the greatest lower bound. I wasn't taught this method before, hence my confusion.

Answer 1

Best that we simplify $\frac{n^2-1}{n^3-1}$ first.

Thus \begin{align} \frac{n^2-1}{n^3-1}& \lt\frac{n^2-1}{n^3-n} \\ &=\frac{n^2-1}{n(n^2-1)} \\ &=\frac{1}{n} \end{align}

We know that $\forall \varepsilon \gt 0, $ $ \exists n_0$ s.t. $n_0 \gt \frac{1}{\varepsilon}$, we have

\begin{align} \frac{n_0^2-1}{n_0^3-1}& \lt \frac{1}{n_0} \\ & \lt \frac{1}{\frac{1}{\varepsilon}} \\ &= \varepsilon \end{align}

Therefore the infimum cannot be a positive number.

i.e. $\inf A \leq 0$

Since $0$ is a lower bound, therefore $\inf A \geq 0$.

Hence $\inf A = 0$.

Answer 2

As I understand the question, the asker is confused about a way of establishing the infimum using an argument involving a limit. If I understand the question correctly, the relevant result is something like the following:

Proposition: Suppose that $A \subseteq \mathbb{R}$, that $m$ is a lower bound of $A$, and that there exists some sequence $(a_n)_{n\in\mathbb{N}} \subseteq A$ with $\lim_{n\to \infty} a_n = m$. Then $\inf A = m$.

Basically, if a real number is both a lower bound of, and a limit point of, some set, then that number is the infimum. This can be proved as follows:

Proof: In the setting of the Proposition, it is sufficient to show that if $b > m$, then there exists some $n$ such that $m \le a_n < b$ (that is, if $b > m$, then $b$ is not a lower bound of $A$). For any $b > m$, choose $\varepsilon \in (0,b-m)$ (for example, choose $\varepsilon = \frac{1}{2}(b-m)$). As $\lim_{n\to\infty} a_n = m$, there exists some $N$ so large that $n > N$ implies that $|m-a_n| < \varepsilon$. But $m$ is a lower bound for $A$ and $a_n \in A$, hence $$ 0 < |m-a_n| = a_n-m < \varepsilon < b-m \implies m \le a_n < b, $$ which completes the proof.

With respect to the question, both $n^2 - 1$ and $n^3 - 1$ are positive for all $n > 1$, which implies that $$ 0 < \frac{n^2-1}{n^3-1}. $$ In other words, $0$ is a lower bound for $$ A = \left\{ \frac{n^2-1}{n^3-1} : n \in \mathbb{N} \right\}. $$ Observe that $$ \lim_{n\to\infty} \frac{n^2 - 1}{n^3 - 1} = \lim_{n\to\infty} \frac{\frac{1}{n} - \frac{1}{n^3}}{1 - \frac{1}{n^3}} = \frac{0-0}{1-0} = 0. $$ Hence $A$ contains a subsequence $a_n = \frac{n^2 -1}{n^3-1}$ with $\lim_{n\to \infty} a_n = 0$. By the Proposition, above, $\inf A = 0$.