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\noindent {\bf Question}

\noindent In this question, $A$ and $B$ are subsets of ${\bf R}$.
Show that each of the following holds.
\begin{enumerate}
\item $\inf(A\cup B) =\min(\inf(A), \inf(B))$;
\item if $A\cap B\ne\emptyset$, then $\sup(A\cap B) \le\min( \sup(A),
\sup(B))$;
\item if $A\cap B\ne\emptyset$, then $\inf(A\cap B)\ge \max( \inf(A),
\inf(B))$;
\item if $u$ is an upper bound for $A$ and if $u\in A$, then $u
=\sup(A)$;
\item if $t$ is an lower bound for $A$ and if $t\in A$, then $t
=\inf(A)$;
\item if $\inf(A)$ exists, then $\inf(A) =\sup\{ y\: |\: y\: {\rm is\:
a \: lower\: bound\: of\: A} \}$;
\item if $\sup(A)$ exists, then $\sup(A) =\inf\{ y\: |\: y\: {\rm is\:
a \: upper\: bound\: of\: A} \}$;
\item $\sup(A)$ is unique if it exists;
\item $\inf(A)$ is unique if it exists;
\end{enumerate}

\medskip
\noindent {\bf Answer}
\begin{enumerate}
\item Assume without loss of generality that $\inf(A)\le\inf(B)$, so
that $\min( \inf(A), \inf(B)) =\inf(A)$.  To show that $\inf(A\cup
B) =\inf(A)$, we need to show two things, that $\inf(A)$ is a
lower bound for $A\cup B$ and that if $t$ is any lower bound for
$A\cup B$, then $t\le\inf(A)$.

\medskip
\noindent If $a\in A$, then $a\ge \inf(A)$ by definition (since
$\inf(A)$ is less than or equal to every element of $A$).
Similarly, if $b\in B$, then $b\ge\inf(B)$; since
$\inf(B)\ge\inf(A)$, this yields that $b\ge \inf(A)$ for all $b\in
B$.  Since every element $c$ of $A\cup B$ satisfies either $c\in
A$ or $c\in B$ (or both), we see that $c\ge \inf(A)$, and so
$\inf(A)$ is a lower bound for $A\cup B$.

\medskip
\noindent Let $t$ be any lower bound for $A\cup B$.  Since $t\le
c$ for every $c\in A\cup B$, we also have that $t\le c$ for every
$c\in A$.  In particular, $t$ is a lower bound for $A$, and so by
the definition of infimum, $t\le\inf(A)$.  Therefore, $\inf(A)$ is
a lower bound for $A\cup B$ that is greater than or equal to any
other lower bound for $A\cup B$.  That is, $\inf(A\cup B)
=\inf(A)$.
\item The easiest way to do this is to begin with an intermediate
fact: if $A\subset B$ and if $\sup(B)$ exists, then $\sup(A)$
exists and $\sup(A)\le \sup(B)$.  The proof uses the definition of
supremum: since $\sup(B)$ exists, we have that $b\le \sup(B)$ for
all $b\in B$ and that if $u$ is an upper bound for $B$, then
$\sup(B)\le u$.  Since $b\le\sup(B)$ for all $b\in B$ and since
$A\subset B$, we have that $a\le \sup(B)$ for all $a\in A$.  In
particular, $A$ is bounded above, and so $\sup(A)$ exists.  To see
the second statement, note that since $\sup(B)$ is an upper bound
for $A$, we have that $\sup(A)\le\sup(B)$ by definition.

\medskip
\noindent So, since $A\cap B\subset A$, we have that $\sup(A\cap
B)\le \sup(A)$.  Similarly, $A\cap B\subset B$, and so $\sup(A\cap
B)\le \sup(B)$.  Hence, $\sup(A\cap B)\le \min(\sup(A),\sup(B))$.

\medskip
\noindent To have an example in which $\sup(A\cap B) <
\min(\sup(A), \sup(B))$, take $A =\{ 0, 1\}$ and $B =\{ 0, 2\}$.
Then, $\sup(A) = 1$, $\sup(B) = 2$, and $\sup(A\cap B) = 0$ since
$A\cap B =\{ 0\}$.
\item The easiest way to do this is to begin with an intermediate
fact: if $A\subset B$ and if $\inf(B)$ exists, then $\inf(A)$
exists and $\inf(A)\ge \inf(B)$.  The proof uses the definition of
infimum: since $\inf(B)$ exists, we have that $b\ge \inf(B)$ for
all $b\in B$ and that if $t$ is a lower bound for $B$, then
$\inf(B)\ge t$.  Since $b\ge\inf(B)$ for all $b\in B$ and since
$A\subset B$, we have that $a\ge \inf(B)$ for all $a\in A$.  In
particular, $A$ is bounded below, and so $\inf(A)$ exists.  To see
the second statement, note that since $\inf(B)$ is a lower bound
for $A$, we have that $\inf(A)\ge\inf(B)$ by definition.

\medskip
\noindent So, since $A\cap B\subset A$, we have that $\inf(A\cap
B)\ge \inf(A)$.  Similarly, $A\cap B\subset B$, and so $\inf(A\cap
B)\ge \inf(B)$.  Hence, $\inf(A\cap B)\ge \max(\inf(A),\inf(B))$.

\medskip
\noindent We note that it is possible to construct an example in
which $\inf(A\cap B) > \max(\inf(A), \inf(B))$.  Namely, take $A
=\{ -1, 0\}$ and $B =\{ -2, 0\}$.  Then, $\inf(A) = -1$, $\inf(B)
= -2$, and $\inf(A\cap B) = 0$ since $A\cap B =\{ 0\}$.
\item Since $u$ is an upper bound for $A$, we have that $u\ge
\sup(A)$, by the definition of supremum.  (And note that $\sup(A)$
exists since $A$ is bounded above.)  Since $u\in A$, we also have
that $u\le \sup(A)$.  Since $u\ge\sup(A)$ and $u\le\sup(A)$, it
must be that $u =\sup(A)$.
\item Since $t$ is a lower bound for $A$, we have that $t\le \inf(A)$,
by the definition of infimum.  (And note that $\inf(A)$ exists
since $A$ is bounded below.)  Since $t\in A$, we also have that
$t\ge \inf(A)$.  Since $t\le\inf(A)$ and $t\ge\inf(A)$, it must be
that $t =\inf(A)$.
\item Set $X =\{ y\: |\: y\mbox{ is a lower bound for A} \}$.  By
definition, $\inf(A)\in X$, since $\inf(A)$ is a lower bound for
$A$. Now take any element $y$ of $X$, so that $y$ is a lower bound
for $A$.  Again by the definition of the infimum, $y\le \inf(A)$.
So, $\inf(A)$ is an upper bound for $X$ and $\inf(A)\in X$, and so
$\inf(A) =\sup(X) =\sup\{ y\: |\: y\mbox{ is a lower bound for A}
\}$.  (Note that the assumption that $\inf(A)$ exists is
equivalent to the assumption that $A$ is bounded below, which
insures that $X$ is non-empty.)
\item Set $X =\{ y\: |\: y\mbox{ is an upper bound for A} \}$.  By
definition, $\sup(A)\in X$, since $\sup(A)$ is an upper bound for
$A$.  Now take any element $y$ of $X$, so that $y$ is an upper
bound for $A$.  Again by the definition of the supremum, $y\ge
\sup(A)$. So, $\sup(A)$ is a lower bound for $X$ and $\sup(A)\in
X$, and so $\sup(A) =\inf(X) =\inf\{ y\: |\: y\mbox{ is an upper
bound for A} \}$.  (Note that the assumption that $\sup(A)$ exists
is equivalent to the assumption that $A$ is bounded above, which
insures that $X$ is non-empty.)
\item This one we argue by contradiction.  Suppose that a set $A$ has
two suprema, and call them $x_1$ and $x_2$.  Both $x_1$ and $x_2$
are upper bounds for $A$, by definition.  Since $x_1$ is a
supremum for $A$, it is less than or equal to all other upper
bounds, and so $x_1\le x_2$.  Similarly, since $x_2$ is a supremum
for $A$, it is less than or equal to all other upper bounds, and
so $x_2\le x_1$. Since $x_1\le x_2\le x_1$, it must be that $x_1
=x_2$, and so the supremum of $A$ is unique.  (Note that this
exercise justifies why we call it 'the supremum' instead of 'a
supremum'.)
\item This one we argue by contradiction.  Suppose that a set $A$ has
two infima, and call them $x_1$ and $x_2$.  Both $x_1$ and $x_2$
are lower bounds for $A$, by definition.  Since $x_1$ is an
infimum for $A$, it is greater than or equal to all other lower
bounds, and so $x_1\ge x_2$.  Similarly, since $x_2$ is an infimum
for $A$, it is greater than or equal to all other upper bounds,
and so $x_2\ge x_1$. Since $x_1\ge x_2\ge x_1$, it must be that
$x_1 =x_2$, and so the infimum of $A$ is unique.  (Note that this
exercise justifies why we call it 'the infimum' instead of 'an
infimum'.)
\end{enumerate}

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