Exercise 3.3.6

Question

This exercise is meant to illustrate the point made in the opening paragraph to Section 3.3. Verify that the following three statements are true if every blank is filled in with the word ``finite.'' Which are true if every blank is filled in with the word ``compact''? Which are true if every blank is filled in with the word ``closed''?
  \begin{enumerate}[label=(\alph*)] 
  \item Every \makebox[1cm]{\hrulefill}\space set has a maximum.
  \item If $A$ and $B$ are \makebox[1cm]{\hrulefill}, then $A+B=\{a+b: a \in A, b \in B\}$ is also \makebox[1cm]{\hrulefill}
  \item If $\left\{A_{n}: n \in \mathbf{N}\right\}$ is a collection of \makebox[1cm]{\hrulefill} sets with the property that every finite subcollection has a nonempty intersection, then $\bigcap_{n=1}^{\infty} A_{n}$ is nonempty as well.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Finite (by taking the maximum value), compact (by taking the supremum, which exists because of boundedness and is in the set because of closed-ness), but not closed ($\mathbf{R}$ is closed)
  \item Finite (by just enumerating through all possibilities of $a + b$).  Compact - boundedness is obviously preserved. For closed-ness, note that if $c \in A+B$ then we can find a convergent sequence $(a_n + b_n) \to c$ in $A + B$. Now since $(a_n)$ is bounded there must be a convergent subsequence which converges to some $a \in A$, and taking the corresponding elements in $(b_n)$ we have a new bounded sequence from which we can get a convergent subsequence which converges to some $b \in B$. Taking the corresponding elements in $(a_n + b_n)$ shows that $c = a+b$. This argument fails if $A$ and $B$ are only closed, as boundedness no longer applies. E.g. for $A = \{n : n \in \mathbf{N}\}$, $B = \{-n + 1/n : n \in \mathbf{N}\} $ then the sequence $(1/n)$ is in $A + B$ which converges to 0, which is not in $A+B$.
  \item Finite - the only way every finite subcollection has a nonempty intersection is if there is at least one element all sets include. Compact, since letting $K_n = \bigcap_{k=1}^n A_{k}$ gives $K_{n} \subseteq K_{n-1}$, we also have $K_n \ne \emptyset$ since every finite intersection is known to be nonempty. Applying the Nested Compact Set Property allows us to conclude
    $$
    \bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty K_n \ne \emptyset
    $$
  Not closed, e.g. let $A_n = [n, \infty)$
   \end{enumerate}

Exercise 3.3.6

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Exercise 3.3.6

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