Exercise 4.3

Question

\begin{enumerate}[label=(\alph*)]
\item $\def\inds{\mathcal{A}}\def\intA{\bigcap_{A \in \inds} A}\def\intB{\bigcap_{B \in \inds} B}$ Show that if $\inds$ is a collection of inductive sets, then the intersection of the elements of $\inds$ is an inductive set.
\item Prove the basic properties (1) and (2) of ${{\mathbb{Z}}_+}$.
\end{enumerate}

Dan Whitman · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item We must show that $\intA$ is inductive.
  \begin{proof}
    First, consider any $A \in \inds$.
    Then, since $A$ is inductive, $1 \in A$.
    Since $A$ was arbitrary, this shows that $1 \in \intA$.
    Now suppose that $x \in \intA$ and again consider arbitrary $A \in \inds$.
    Then $x \in A$ so that $x+1 \in A$ also since $A$ is inductive.
    Since $A$ was arbitrary, this shows that $x+1 \in \intA$.
    Hence by definition $\intA$ is inductive.
  \end{proof}

\item
  \begin{proof}
    Let $\inds$ be the collection of all inductive sets of ${\mathbb{R}}$ so that by definition ${{\mathbb{Z}}_+} = \intA$.
    It then follows immediately from part~(a) that ${{\mathbb{Z}}_+}$ is inductive since $\inds$ is a collection of inductive sets.
    This shows property (1).

Now suppose that $A$ is an inductive set of positive integers.
    That is, $A$ is inductive and $A \subset {{\mathbb{Z}}_+}$.
    Consider any $x \in {{\mathbb{Z}}_+} = \intB$, where again $\inds$ is the the collection of all inductive subsets of ${\mathbb{R}}$.
    Clearly we have that $A \subset {{\mathbb{Z}}_+} \subset {\mathbb{R}}$ so that $A \in \inds$ since $A$ is an inductive subset of ${\mathbb{R}}$.
    Hence $x \in A$ (since $x \in \intB$ and $A \in \inds$) so that ${{\mathbb{Z}}_+} \subset A$ since $x$ was arbitrary.
    This shows that $A = {{\mathbb{Z}}_+}$ as desired since also $A \subset {{\mathbb{Z}}_+}$.
    This shows property (2).
  \end{proof}
\end{enumerate}

Exercise 4.3

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Exercise 4.3

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