Exercise 1.2.3

Question

Decide which of the following represent true statements about the nature of sets. For any that are false, provide a specific example where the statement in question does not hold.
  \begin{enumerate}[label=(\alph*)] 
  \item If $A_{1} \supseteq A_{2} \supseteq A_{3} \supseteq A_{4} \cdots$ are all sets containing an infinite number of elements, then the intersection $\bigcap_{n=1}^{\infty} A_{n}$ is infinite as well.
  \item If $A_{1} \supseteq A_{2} \supseteq A_{3} \supseteq A_{4} \cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n=1}^{\infty} A_{n}$ is finite and nonempty.
  \item $A \cap(B \cup C)=(A \cap B) \cup C$.
  \item $A \cap(B \cap C)=(A \cap B) \cap C$.
  \item $A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item False, consider $A_1 = \{1, 2, \dots\}, A_2 = \{2, 3, \dots\}$, ... has $\bigcap_{n=1}^\infty A_n = \emptyset$.
  \item True, because we eventually reach $A_j = \{x\}$ and get stuck
  \item False, $A = \emptyset$ gives $\emptyset = C$.
  \item True, intersection is associative.
  \item True, draw a diagram.
   \end{enumerate}

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} 
b) In Abbott, $\N = \{1,2,3,\ldots\}$.

We start from the property of $\N$ which says that every non empty subset $E$ of $\N$ has a least element, denoted by $\min(E)$.

Consider the set $E$ of the finite cardinalities of the sets $A_1,A_2,A_3,\ldots$
$$E = \{ j  \in \N \mid, \exists i \in \N,\ j = |A_i|\},$$
or, more biefly,
$$E = \{|A_i|,\  i\in \N\}.$$
Since every $A_i$ is non empty, $|A_i| \geq 1$, and every $A_i$ is finite, so $E \subseteq \N$ (and $|A_1| \in E$, so $E 
e \varnothing$). Therefore $E$ has a least  element
$$n = \min \{|A_i|,\ i \in \N\}.$$
Since $n \in E$ (by definition of a least element), $n = |A_k|$ for some index $k \in \N$, and, for every index $i\in \N$,
\begin{align}
n = |A_k| \leq |A_i|.
\end{align}
By hypothesis, if $j\geq k$, then $A_j \subseteq A_k$, thus $|A_j| \leq |A_k|$. But, from (1), we have also $|A_k| \leq |A_j|$, therefore $|A_j| = |A_k|$.

From $|A_j| = |A_k|$ and $A_j \subseteq A_k$, where $A_j,A_k$ are finite, we can deduce $A_j = A_k$. This gives
$$\forall j \geq k, \ A_j = A_k.$$
(We say that the sequence $(A_i)_{i\in \N}$ is stationary.)

Hence
$$\bigcap_{i \in N} A_i =\bigcap_{1\leq i \leq k} A_i \cap \bigcap_{i>k} A_j = A_k \cap A_k = A_k.$$
Since $A_k$ is finite, nonempty, $\bigcap\limits_{i \in N} A_i = A_k$ is finite, nonempty. 
\end{proof}

Exercise 1.2.3

Answers

Comments

Comments

Exercise 1.2.3

Answers

Comments

Comments

Add answer