Compute the supremum and the infimum of the following sets
- \(\{n \in \mathbb{N}: n^2 < 10\}\)
- \(\{n/(m+n): m,n \in \mathbb{N}\}\)
- \(\{n/(2n+1): n \in \mathbb{N}\}\)
- \(\{n/m: m,n \in \mathbb{N}\}\) with \(m + n \leq 10\)
Solution
- \(A = \{n \in \mathbb{N}: n^2 < 10\}\).
In this case, \(n \leq \sqrt{10}\). So \(\sup A = 3\) while \(\inf A = 1\).
- \(A = \{n/(m+n): m,n \in \mathbb{N}\}\).
Note here that$$ \begin{align*} 0 < \frac{n}{m+n} < 1. \end{align*} $$So \(1 \not\in A\) and \(0 \not\in A\) but many values exist in between. Therefore, \(\sup A = 1\) while \(\inf A = 0\).
- \(A = \{n/(2n+1): n \in \mathbb{N}\}\)
Take \(n = 1\), then \(1/(2n+1) = 1/3\).
Now, as \(n \rightarrow \infty\), then$$ \begin{align*} \lim_{n \rightarrow \infty} \frac{n}{2n+1} &= \lim_{n \rightarrow \infty} \frac{n}{n(2 + \frac{1}{n})} \\ &= \lim_{n \rightarrow \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2} \end{align*} $$Therefore, \(\sup A = \frac{1}{2}\) and \(\inf A = \frac{1}{3}\).
- \(A = \{n/m: m,n \in \mathbb{N}\}\) with \(m + n \leq 10\)
Take \(m = 1\) and \(n = 9\) so \(n/m=9\). Therefore, \(\sup A = 9\).
Take \(n = 1\) and \(m = 9\) so \(n/m=1/9\). Therefore, \(\inf A = 1/9\).
References