I am self-studying Real Analysis from the text, Understanding Analysis by Stephen Abbott. I would like someone to verify my justifications/counterexamples to the following assertions about closed and compact sets. This is exercise 3.3.6.
This exercise is meant to illustrate the point made in the opening paragraph to the section 3.3. Verify that the following three statements are true if every blank is filled 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?
(a) Every __________ set has a maximum.
(i) Finite.
We use the principle of mathematical induction. Let $\displaystyle a_{k} \in \mathbf{R}$. Consider the singleton set
\begin{equation*}
A_{1} =\{a_{1}\}
\end{equation*}
Clearly $\displaystyle a_{1}$ is the maximal element since.
Consider a set consisting of a pair of elements.
\begin{equation*}
A_{2} =\{a_{1} ,a_{2}\}
\end{equation*}
By the trichotomy property, atleast one of the three, but not all are true
(1) $\displaystyle a_{1} < a_{2}$
(2) $\displaystyle a_{1} =a_{2}$
(3) $\displaystyle a_{1} >a_{2}$.
In any of these cases, $\displaystyle A_{2}$ has a maximum element such that
\begin{equation*}
\begin{array}{ c c }
a_{1} & \leq \max\{a_{1} ,a_{2}\}\\
a_{2} & \leq \max\{a_{1} ,a_{2}\}
\end{array}
\end{equation*}
Assume that $\displaystyle A_{n} =\{a_{1} ,a_{2} ,\dotsc ,a_{n}\}$ has a maximum.
We are interested to prove that $\displaystyle A_{n+1} =\{a_{1} ,a_{2} ,\dotsc ,a_{n+1}\}$ has a maximum.
\begin{equation*}
\max\{a_{1} ,a_{2} ,\dotsc ,a_{n} ,a_{n+1}\} =\max\{\max A_{n} ,a_{n+1}\}
\end{equation*}
Since, $\displaystyle \max A_{n}$ is well defined, we can compare it with $\displaystyle a_{n+1}$ to determine, the largest element. Consequently, $\displaystyle A_{n+1}$ has a largest element.
Thus, $\displaystyle A_{n}$ has a maximum for all $\displaystyle n\in \mathbf{N}$.
(ii) Compact.
This proposition is true.
Let $K$ be a compact set. By the theorem on characterization of compact sets, if $K$ is compact, then $K$ is closed and bounded. Since $K$ is bounded subset of $\mathbf{R}$, by the least upper bound property, $K$ has a least upper bound.
Let $s=\sup K$. By definition, for all $\epsilon>0$, there exists $a_n \in K$, such that $s-\epsilon < a_n < s < s + \epsilon$. Consequently, we have produced a sequence $(a_n) \subseteq K$, such that $\lim a_n = s$. So, $s$ is a limit point of $K$. Since, $K$ is closed, $s \in K$. Thus, $K$ has a maximal element.
(iii) Closed.
This proposition is false. $\mathbf{R}$ is closed, but has no maximal element.
(b) If $\displaystyle A$ and $\displaystyle B$ are ________, then $\displaystyle A+B=\{a+b:a\in A,b\in B\}$ is also ________.
(i) Finite.
The cartesian product of finite sets $\displaystyle A$ and $\displaystyle B$, $\displaystyle A\times B$ is finite. Since, $\displaystyle +:A\times B\rightarrow A+B$ is a well-defined function on a finite set, each element in the domain, maps to one and only one element in the co-domain. So, the image set $\displaystyle A+B$ is also finite.
(ii) Compact.
This proposition is true.
We are interested to prove that $A+B$ is compact. Let $(c_n)$ be an arbitrary sequence in $A+B$. By definition of the set $A+B$, $c_n = a_n + b_n$, where $a_n \in A$, $b_n \in B$. So, the terms of every sequence in $A+B$, is the sum of the terms of a sequence contained in $A$, and another sequence contained in $B$.
As $A$ and $B$ are compact sets, the sequence $(a_n)$ has a subsequence $(a_{n_k})$, which converges to a limit $\lim a_{n_k} = a$ which is also in $A$; the sequence $(b_n)$ has a subsequence $(b_{n_k})$, which converges to a limit $\lim b_{n_k} = b$ which is also in $B$. By definition of $A+B$, the elements $a_{n_k} + b_{n_k}$ are contained in $A+B$ and is subsequence of $(a_n) + (b_n)$. By the Algebraic Limit Theorem, $\lim (a_{n_k} + b_{n_k}) = \lim (a_{n_k}) + \lim (b_{n_k}) = a + b$ which belongs to $A+B$. So, $A+B$ is compact.
(iii) Closed.
Not sure, how to go about this.
(c) If $\displaystyle \{A_{n} :n\in \mathbf{N}\}$ is a collection of _________ sets with the property that every finite subcollection has a non-empty intersection, then $\displaystyle \bigcap _{n=1}^{\infty } A_{n}$ is non-empty as well.
(i) Finite.
Assume that $\bigcap_{n=1}^{\infty}A_n$ is empty. Then, there must be atleast one finite subcollection of $\{A_n:n \in \mathbf{N}\}$ which does not have a element. But, this contradicts the fact that, every finite subcollection is non-empty. Consequently, $\bigcap_{n=1}^{\infty}A_n \neq \emptyset$.
(ii) Compact.
Let
\begin{align*}
A_1' &= A_1\\
A_2' &= A_1 \cap A_2 \\
A_3' &= A_1 \cap A_2 \cap A_3 \\
\vdots\\
A_n' &= \bigcap_{k=1}^{n} A_k
\end{align*}
Since, the arbitrary intersection of compact sets is compact, $A_1', A_2', \ldots$ are non-empty compact sets. We have sequence of non-empty nested compact sets –
$$A_1' \supseteq A_2' \supseteq A_3' \supseteq \ldots$$
By the nested compact set property, their intersection is non-empty.
(iii) Closed.
This proposition is false, because a nested sequence of non-empty closed sets do not always have a non-empty intersection.
Best Answer
In order for $A+B$ not to be closed when $A$ and $B$ are closed, the hint is to choose sets $A$ and $B$ such that at least one of them is infinite. A better way is to construct two divergent sequences of isolated points $\{a_n\}$ and $\{b_n\}$, such that $a_n+b n$ converges to a value $a$, which is not equal to $a_n+b_m$ for any $n$ and $m$. I could guess one: $$ A= \{n+\frac{1}{n}| n\in \mathbb{N}, n>1\} \quad \text{and} \quad B= \{-n |n\in \mathbb{N}\}.$$
The rest seem to be correct.