Solution: Let $A$ be a nonempty subset of $\mathbb{R} $.

1. Upper bound: We say that $\alpha $ is an upper bound of $A$ if for every $a\in A,~a\leq \alpha $. In terms of quantifiers \[ \forall a\in A,~a\leq \alpha. \]
2. Lower bound: We say that $\beta $ is a lower bound of $A$ if for every $a\in A,~\beta \leq a$. In terms of quantifiers \[ \forall a\in A,~\beta \leq a. \]
3. Least upper bound: We say that $\alpha$ is a least upper bound of $A$ if
   1. $\alpha $ is an upper bound of $A$ and
   2. If $\beta $ is an upper bound of $A$, then $\alpha \leq \beta $ ( that is, least among all upper bounds).
   
   There is an equivalent definition in terms of $\varepsilon $. \[ \forall~ \varepsilon >0, ~\exists ~a \in A \text{ such that } \alpha -\varepsilon < a. \]
4. Greatest lower bound: We say that $\beta$ is a greatest lower bound of $A$ if
   1. $\beta $ is a lower bound of $A$ and
   2. If $\alpha $ is a lower bound of $A$, then $\alpha \leq \beta $ ( that is, greatest among all lower bounds).
   
   There is an equivalent definition in terms of $\varepsilon $. \[ \forall~ \varepsilon >0, ~\exists ~a \in A \text{ such that } a < \beta + \varepsilon. \]

1. Given that $\alpha $ is a least upper bound of $A$ and $\beta $ is a greatest lower bound of $A$. We need to show that $\beta \leq \alpha $. Note that for any $a\in A$ we have \[ \beta \leq a \leq \alpha \implies \beta \leq \alpha . \]
2. At first we will show that LUB of $A$ is unique. Let $\alpha $ and $\alpha ^\prime $ be two LUBs of $A$. Since $\alpha $ is an LUB and $\alpha ^\prime$ is an upper bound so we must have \begin{equation}\label{eq:06Mar2023-1} \alpha \leq \alpha ^\prime . \end{equation} Similarly, $\alpha ^\prime $ is an LUB and $\alpha $ is an upper bound implies \begin{equation}\label{eq:06Mar2023-2} \alpha ^\prime \leq \alpha . \end{equation} Using, \eqref{eq:06Mar2023-1} and \eqref{eq:06Mar2023-2}, we conclude that $\alpha =\alpha ^\prime $ and hence we can use ‘the’ LUB of $A$.

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   Now we will show that GLB of $A$ is unique. Let $\beta $ and $\beta ^\prime $ be two GLBs of $A$. Since $\beta $ is an GLB and $\beta ^\prime$ is a lower bound so we must have \begin{equation}\label{eq:06Mar2023-3} \beta^\prime \leq \beta. \end{equation} Similarly, $\beta ^\prime $ is an GLB and $\beta $ is a lower bound implies \begin{equation}\label{eq:06Mar2023-4} \beta \leq \beta^\prime. \end{equation} Using, \eqref{eq:06Mar2023-3} and \eqref{eq:06Mar2023-4}, we conclude that $\beta =\beta ^\prime $ and hence we can use ‘the’ GLB of $A$.