Find the supremum and the infimum of \begin{equation*} A = \left\{\left(-\frac{1}{m}, \frac{1}{n}\right): m,n \in \Bbb N\right\} \end{equation*} on the set $X = 2^{\Bbb R}$ (the set of all subset of $\Bbb R$) which is partially ordered by inclusion. (Note: all element in $A$ is an open interval).
Definition. Let $S$ be a non-empty set. A set $X=2^S$ (the set of all subset of $S$) with partially ordered inclusion: \begin{equation*} A \le B \iff A \subseteq B, \quad \forall A,B \in X. \end{equation*}
Attempt: I claim that $\sup A = (-1,1)$ and $\inf A$ doesn't exist. Firstly, we'll show that $\sup A = (-1,1)$. Notice that for all $m,n \in \Bbb N$, we have $-1 \le -\frac{1}{m} < 0$ and $0 < \frac{1}{n} \le 1$. Hence, $\left(-\frac{1}{m}, \frac{1}{n}\right) \subseteq (-1,1)$. By definition above, we have $\left(-\frac{1}{m}, \frac{1}{n}\right) \le (-1,1)$. Therefore, $(-1,1)$ is an upper bound of $A$. Now, let $(a,b) \in X$ be another upper bound of $A$. Then, \begin{equation*} \left(-\frac{1}{m}, \frac{1}{n}\right) \le (a,b) \quad \forall m,n \in \Bbb N \iff \left(-\frac{1}{m}, \frac{1}{n}\right) \subseteq (a,b), \forall m,n \in \Bbb N. \qquad (\star) \end{equation*} Now, claim that $a \le -1$ and $1 \le b$. Suppose that $a > -1$ and $b < 1$. Then, $(a,b) \subseteq (-1,1)$, and so $(a,b) \le (-1,1)$. This contradicts $(\star)$. Hence, $a \le -1$ and $1 \le b$. Therefore, \begin{equation*} (-1,1) \subseteq (a,b) \iff (-1,1) \le (a,b). \end{equation*} Thus, $(-1,1)$ is the least upper bound of $A$. Hence, $\sup A = (-1,1)$.
Now, suppose that $\inf A$ exists, namely, $(a,b)$. Then, \begin{equation*} (a,b) \le \left(-\frac{1}{m}, \frac{1}{n}\right), \forall m,n \in \Bbb N \iff (a,b) \subseteq \left(-\frac{1}{m}, \frac{1}{n}\right), \forall m,n \in \Bbb N. \end{equation*} That means that $-\frac{1}{m} \le a < 0$ and $0 < b \le \frac{1}{n}$, for all $m,n \in \Bbb N$. So, the value of $a$ and $b$ are depending on the value of $m$ and $n$, respectively, i.e., $(a,b)$ is not unique. This contradicts the fact that the infimum of a set (if exists) is unique. Thus, $\inf A$ doesn't exist.
Does the above proof correct? Thanks in advanced.
First of all, you want to find the infimum and supremum in $\mathcal{P}(\mathbb{R})$, not in $A$, so the infimum does exist and is $\{0\}$, because $\{0\}$ and $\varnothing$ are the only sets which are subsets of all elements of $A$[Since $\bigcap A=\{0\}$]. The supremum is $(-1,1)$ because: it’s an upper bound, and if $M$ is an upper bound then $(-1,1)$ must be a subset of $M$ since $(-1,1)\in A$.