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\title{On minimums and maximums}

\author{Branko \'{C}urgus}


\begin{document}


\maketitle


\begin{definition} \label{de:sub} 
Let $A$ and $B$ be two sets. We say that $A$ is a {\em subset} of $B$, and write $A\subseteq B$, if and only if for every $x \in A$ we have $x \in B$. In notation 
\begin{equation} \label{eq:sub}
A \subseteq B \qquad \Leftrightarrow \qquad \forall\mkern+0.5mu x \ \bigl( x \in A \Rightarrow x \in B  \bigr),
\end{equation}
\end{definition}


\begin{definition} \label{de:minmax}
Let $S$ be a subset of $\mathbb{R}$. If $u$ is the smallest number in $S$, then $u$  is called a {\it minimum} of $S$ and we write $u = \min S$.  If $v$ is the greatest number in $S$, then $v$ is called a {\it maximum} of $S$ and we write $v = \max S$. More formally, we express these definitions as logical statements:
\begin{equation} \label{eq:min}
u = \min S \qquad \Leftrightarrow \qquad (u \in S) \land  \bigl(\forall \mkern+0.5mu x \in S \ \ u \leq x  \bigr),
\end{equation}
\begin{equation} \label{eq:max}
v = \max S \qquad \Leftrightarrow \qquad (v \in S) \land  \bigl(\forall \mkern+0.5mu x \in S \ \ x \leq v  \bigr).
\end{equation}
\end{definition}


\begin{proposition} \label{pr:mmsub}
Let $A$ and $B$ be nonempty subsets of $\mathbb{R}$ such that $A\subseteq B$. The following statements hold.  
\begin{enumerate}
\renewcommand*\theenumi{\roman{enumi}}
\renewcommand*\labelenumi{{\rm (\theenumi)}}
\item \label{pr:mmsub-i1}
If the sets $A$ and $B$ have minimums, then
\begin{equation}\label{eq:minsub}
\min B \leq \min A. 
\end{equation}

\item \label{pr:mmsub-i2}
If the sets $A$ and $B$ have maximums, then
\begin{equation}\label{eq:maxsub}
\max A \leq \max B.
\end{equation}
\end{enumerate}
\end{proposition}
\begin{proof}
Assume $A$ and $B$ be nonempty subsets of $\mathbb{R}$ such that $A\subseteq B$. 

(\ref{pr:mmsub-i1}) Assume that $A$ and $B$ have minimums and set $a = \min A$ and $b = \min B$. 
By definition of the minimum, see $\Rightarrow$ in \eqref{eq:min}, we have $a \in A$. By definition of the subset, see $\Rightarrow$ in \eqref{eq:sub}, $a\in A$ implies $a\in B$. Hence $a \in B$ holds.  Since $b = \min B$, by definition of the minimum, see $\Rightarrow$ in \eqref{eq:min}, we have that $b \leq y$ for all $y \in B$. Since we already proved that $a \in B$, we conclude that $b \leq a$. This proves \eqref{eq:minsub}.


(\ref{pr:mmsub-i2}) Assume that $A$ and $B$ have maximums and set $c = \max A$ and $d = \max B$.
By definition of the maximum, see $\Rightarrow$ in \eqref{eq:max}, we have $c \in A$. By definition of the subset, see $\Rightarrow$ in \eqref{eq:sub}, we deduce that $c\in B$. Since $d = \max B$, by definition of the maximum, see $\Rightarrow$ in \eqref{eq:max}, we have that $y \leq d$ for all $y \in B$. Since we already proved that $c \in B$, we conclude that $c \leq d$. This proves \eqref{eq:maxsub}.
\end{proof}


\begin{definition} \label{de:sum}
Let $A$ and $B$ be nonempty subsets of $\mathbb{R}$.  We define the {\em sum} $A+B$ to be the following set
\begin{equation} \label{eq:sum} 
A+B = \bigl\{ x+y \in \mathbb{R} : (x\in A) \land (y\in B) \bigr\}. 
\end{equation}
\end{definition}


\begin{proposition} \label{pr:mm-sum}
Let $A$ and $B$ be nonempty subsets of $\mathbb{R}$. The following statements hold.
\begin{enumerate}
\renewcommand*\theenumi{\roman{enumi}}
\renewcommand*\labelenumi{{\rm (\theenumi)}}
\item \label{pr:mm-sum-i1}
If the sets $A$ and $B$ have minimums, then
\begin{equation}\label{eq:minsum}
\min(A+B) = \min A + \min B.
\end{equation}

\item \label{pr:mm-sum-i2}
If the sets $A$ and $B$ have maximums, then
\begin{equation}\label{eq:maxsum}
\max(A+B) = \max A + \max B.
\end{equation}
\end{enumerate}
\end{proposition}

\begin{proof}
Assume $A$ and $B$ be nonempty subsets of $\mathbb{R}$. 

(\ref{pr:mm-sum-i1}) Assume that $A$ and $B$ have minimums and set $a = \min A$ and $b = \min B$.
By definition of the minimum, see $\Rightarrow$ in \eqref{eq:min}, we have that $a \in A$ and $b \in B$. By definition of the sum for two sets, see \eqref{eq:sum}, we have $a+b\in A+B$. By definition of the minimum, see $\Rightarrow$ in \eqref{eq:min}, we have that $z \geq \min(A+B)$ for all $z \in A+B$. Since we already proved that $a+b \in A+B$, we conclude that $a+b \geq \min(A+B)$. This proves \eqref{eq:minsum}.


(\ref{pr:mm-sum-i2}) Assume that $A$ and $B$ have maximums and set $c = \max A$ and $d = \max B$.
By definition of the maximum, see $\Rightarrow$ in \eqref{eq:max}, we have that $a \in A$ and $b \in B$. By definition of the sum for two sets, see \eqref{eq:sum}, we have $a+b\in A+B$. By definition of the maximum, see $\Rightarrow$ in \eqref{eq:max}, we have that $z \leq \max(A+B)$ for all $z \in A+B$. Since we already proved that $a+b \in A+B$, we conclude that $a+b \leq \min(A+B)$. This proves \eqref{eq:maxsum}.
\end{proof}


\end{document}
