Exercise 5.3

Question

Let $A = A_1 	imes A_2 	imes \cdots$ and $B = B_1 	imes B_2 	imes \cdots$.
\begin{enumerate}[label=(\alph*)]
\item Show that if $B_i \subset A_i$ for all $i$, then $B \subset A$.
  (Strictly speaking, if we are given a function mapping the index set ${{\mathbb{Z}}_+}$ into the union of the sets $B_i$, we must change its range before it can be considered as a function mapping ${{\mathbb{Z}}_+}$ into the union of the sets $A_i$.
  We shall ignore this technicality when dealing with cartesian products)
\item Show the converse of (a) holds if $B$ is nonempty.
\item Show that if $A$ is nonempty, each $A_i$ is nonempty.
  Does the converse hold?
  (We will return to this question in the exercises of \S 19.)
\item What is the relation between the set $A \cup B$ and the cartesian product of the sets $A_i \cup B_i$?
  What is the relation between the set $A \cap B$ and the cartesian product of the sets $A_i \cap B_i$?
\end{enumerate}

Dan Whitman · Accepted Answer

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}}

ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}}

ewcommand\vect[1]{\mathbf{#1}}
\def\vw{\vect{w}}
\def\vx{\vect{x}}
\def\vy{\vect{y}}
\def\vz{\vect{z}}
\def\va{\vect{a}}
\def\vb{\vect{b}}
\def\vze{\vect{0}}
\begin{enumerate}[label=(\alph*)]
  \item
  \begin{proof}
    Suppose that $\vb \in B$ so that $\vb = \seqinf{b}$ where $b_i \in B_i$ for every $i \in {{\mathbb{Z}}_+}$.
    Consider any such $i \in {{\mathbb{Z}}_+}$ so that $b_i \in B_i$.
    Then also $b_i \in A_i$ since $B_i \subset A_i$.
    Since $i$ was arbitrary, $b_i \in A_i$ for every $i \in {{\mathbb{Z}}_+}$ so that $\vb = \seqinf{b} \in A_1 	imes A_2 	imes \cdots = A$.
    Since $\vb$ was arbitrary, this shows that $B \subset A$.
    Note that we ignore the function range technicality issue mentioned above.
  \end{proof}

\item
  \begin{proof}
    Suppose that $B \subset A$.
    Since $B 
eq \varnothing$, there is a $\vb' \in B$ so that $\vb' = \seqinf{b'}$ where $b_i' \in B_i$ for every $i \in {{\mathbb{Z}}_+}$.
    Now consider any $i \in {{\mathbb{Z}}_+}$ and $b_0 \in B_i$.
    Then define
    \begin{gather*}
      b_j = \begin{cases}
        b_0 & j = i \
        b_j' & j 
eq i
      \end{cases}
    \end{gather*}
    for any $j \in {{\mathbb{Z}}_+}$.
    Clearly we have that $b_j \in B_j$ for any $j \in {{\mathbb{Z}}_+}$ so that $\vb = \seqinf{b} \in B_1 	imes B_2 	imes \cdots = B$.
    It then follows that also $\vb \in A$ since $B \subset A$.
    Hence $b_j \in A_j$ for every $j \in {{\mathbb{Z}}_+}$.
    In particular, we have $b_0 = b_i \in A_i$.
    Since $b_0$ was arbitrary, this shows that $B_i \subset A_i$, and since $i$ was arbitrary, this shows the desired result.
  \end{proof}

\item
  \begin{proof}
    Suppose that $A$ is nonempty so that there is an $\va \in A$.
    Then, since $A = A_1 	imes A_2 	imes \cdots$, it follows that $\va = \seqinf{a}$ where $a_i \in A_i$ for every $i \in {{\mathbb{Z}}_+}$.
    Therefore, for any such $i \in {{\mathbb{Z}}_+}$, we have that $a_i \in A_i$ so that $A_i 
eq \varnothing$.
    Hence every $A_i$ is nonempty as desired since $i$ was arbitrary.
  \end{proof}

Consider the converse.
  Suppose that each $A_i$ is nonempty (for $i \in {{\mathbb{Z}}_+}$).
  Then there is an $a_i \in A_i$ for every $i \in {{\mathbb{Z}}_+}$ so that $\va = \seqinf{a} \in A_1 	imes A_2 	imes \cdots = A$ so that then $A 
eq \varnothing$.
  While this may seem like an innocuous argument, especially out of the context of axiomatic set theory, it actually requires the Axiom of Choice.
  The reason is that, in the general case when each $A_i$ may have more than one element, or even an infinite number of elements, we have to choose a specific $a_i$ in each $A_i$.
  Since the index set ${{\mathbb{Z}}_+}$ is infinite, an infinite number of these choices must be made, which is precisely when the Axiom of Choice is required.
  If the index set was finite, then the axiom would not be needed.

\item
  First, let $C_i = A_i \cup B_i$ for every $i \in {{\mathbb{Z}}_+}$, and let $C = C_1 	imes C_2 	imes \cdots$,  so that we are asked to compare $C$ with $A \cup B$.

We claim that $A \cup B \subset C$ but that $C$ is \emph{not} generally a subset of $A \cup B$.
  
  \begin{proof}
    First consider any $\vx \in A \cup B$ so that $\vx \in A$ or $x \in B$.
    If $\vx \in A$ then it has to be that $\vx = \seqinf{x}$ where $x_i \in A_i$ for every $i \in {{\mathbb{Z}}_+}$.
    Consider then any such $i \in {{\mathbb{Z}}_+}$.
    Then $x_i \in A_i$ so that clearly $x_i \in A_i \cup B_i = C_i$.
    Since $i$ was arbitrary, we conclude that $\vx = \seqinf{x} \in C_1 	imes C_2 	imes \cdots = C$.
    An analogous argument shows that $\vx \in C$ when $\vx \in B$ as well.
    Hence $A \cup B \subset C$ since $\vx$ was arbitrary.

To show that $C$ is \emph{not} a subset of $A \cup B$ in general, consider the following counterexample.
    Let $A_1 = \varnothing$ and $A_i = \left\{1ight\}$ for every $i \in {{\mathbb{Z}}_+}$ where $i > 1$.
    Also let $B_i = \left\{2ight\}$ for every $i \in {{\mathbb{Z}}_+}$.
    Now, it follows from the contrapositive of part~(c) that $A = \varnothing$ since $A_1 = \varnothing$.
    We also clearly have $B = B_1 	imes B_2 	imes \cdots = \left\{(2, 2, \ldots)ight\}$ so that $A \cup B = \varnothing \cup B = B = \left\{(2, 2, \ldots)ight\}$.
    Clearly $C_1 = A_1 \cup B_1 = \varnothing \cup \left\{2ight\} = \left\{2ight\}$ while, for $i > 1$ we have $C_i = A_i \cup B_i = \left\{1ight\} \cup \left\{2ight\} = \left\{1,2ight\}$.
    It then follows that, for $a_1 = 2$ and $a_i = 1$ for $i > 1$, we have $\va = \seqinf{a} = (2, 1, 1, \ldots) \in C_1 	imes C_2 	imes \cdots = C$.
    However, clearly $\va 
otin A \cup B$, which suffices to show that $C$ cannot be a subset of $A \cup B$ in general.
  \end{proof}

Now let $C_i = A_i \cap B_i$ for every $i \in {{\mathbb{Z}}_+}$ so that we are asked to compare $C = C_1 	imes C_2 	imes \cdots$ and $A \cap B$.

Here we claim that in fact $A \cap B = C$.

\begin{proof}
    First consider any $\vx \in A \cap B$ so that $\vx \in A$ and $\vx \in B$.
    It then follows that $\vx = \seqinf{x}$ where $x_i \in A_i$ for every $i \in {{\mathbb{Z}}_+}$ and $x_i \in B_i$ for every $i \in {{\mathbb{Z}}_+}$.
    Then, for any such $i \in {{\mathbb{Z}}_+}$, clearly $x_i \in A_i$ and $x_i \in B_i$ so that $x_i \in A_i \cap B_i = C_i$.
    We then have that $\vx = \seqinf{x} \in C_1 	imes C_2 	imes \cdots = C$.
    Since $\vx$ was arbitrary, this shows that $A \cap B \subset C$.

Now consider any $\vx \in C$ so that $\vx = \seqinf{x}$ where $x_i \in C_i$ for any $i \in {{\mathbb{Z}}_+}$.
    Then, for any such $i \in {{\mathbb{Z}}_+}$, we have $x_i \in C_i = A_i \cap B_i$ so that $x_i \in A_i$ and $x_i \in B_i$.
    Since $i$ was arbitrary, this shows that both $\vx = \seqinf{x} \in A_1 	imes A_2 	imes \cdots = A$ and $\vx = \seqinf{x} \in B_1 	imes B_2 	imes \cdots = B$.
    Hence $\vx \in A \cap B$, which shows that $C \subset A \cap B$ since $\vx$ was arbitrary.

Therefore it must be that $A \cap B = C$ as desired.
  \end{proof}
\end{enumerate}

Exercise 5.3

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Exercise 5.3

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