Exercise 2.13

Question

\def\ints{\mathbb{Z}}
Exercise 13: Construct a compact set of real numbers whose limit points form a countable set.

ghostofgarborg · Accepted Answer

\def\ints{\mathbb{Z}}

Note that $\left| \frac 1 n - \frac {1}{n-1} \right| = \frac 1 {n(n-1)} > \frac 1 {n^2}$, so that the sets
    $$ S_n = \{ \frac 1 n + \frac 1 m : m \geq n^2 \} \cup \{ \frac 1 n \} $$
    are disjoint. Consider the set $ S = \bigcup_n S_n \cup \{ 0 \} $. The set is clearly bounded, so 
    we need to show that it is closed, and that the limit points are countable.

We claim that the limit points are precisely $ L = \{ \frac 1 n : n \in \mathbb N \} \cup {0} $. 
    Since $L$ is contained in $S$ and is countable, that will suffice to prove our claim. It is easy to 
    see that these points are limit points. To see that there are no other, observe that 
    any point not in $[0,2]$ or the previously mentioned set
    is either strictly greater than the lub or strictly smaller than the glb of all but at most one 
    of the disjoint sets $S_j$.
    It would consequently have to be a limit point of $S_j$, but all such limit points
    are contained in $L$. Lastly, we can easily see that $0$ is a limit point, but $2$ is not.

Amit Awasthi · Answer

\begin{theorem}[Sequential Compactness]
      If the metric space $X$ is separable, then $K \subset X$ is compact if and only if every sequence $\{x_n\} \subset K$ has a convergent subsequence that converges to an element of $K$. This property will be referred to as ``sequential compactness.''
    \end{theorem}
    \begin{proof}
      First prove the compactness $K$ implies sequential compactness. Suppose not. Then, for every $k \in K$ we can find an $r_k > 0$ such that the ball centered at $k$ of radius $r_k$, $B(k,r_k)$, has only finitely many $x_n$'s. These balls form an open cover of $K$, and by the compactness of $K$, there is a finite subcover. This implies $\{x_n\}$ is a finite sequence, which is a contradiction.
      \par Now prove sequential compactness implies compactness. There is a countable open cover of $K$, that is, $K \subset \bigcup_{j=1}^\infty U_j$, which is possible because $X$ is assumed to be separable, and therefore we can construct an open set $U_j$ which is a ball around every member of the countable subset of $K$ with radius $\delta > 0$, whose union forms an open cover of $K$. Suppose, to get a contradiction, that for every $N$, $K 
subseteq \bigcup_{j=1}^N U_j$, which implies for every $N$ there is an $x_n \in K$ such that $x_n 
otin \bigcup_{j=1}^N U_j$; $\{x_n\}$ forms an infinite sequence in $K$. By sequential compactness, there is a subsequence of this sequence $\{x_{nj}\}$ such that $x_{nj} 	o x \in K$ as $j 	o \infty$. This implies $x$ is a member of some set $U_{j0}$ which is part of the open cover of $K$. By the definition of convergence, this implies there is some $M$ such that $|x_{nm} - x| < \delta$ for $m > M$ with $\delta$ as given above, and therefore $x_{nm} \in U_{j0}$ for $m > M$. This is a contradiction, since $\{x_n\}$ was constructed such that all $x_n 
otin \bigcup_{j=1}^N U_j$.
    \end{proof}
    \begin{claim}
      Construct our set $X'$ as follows. First, let $X$ be defined as
      \begin{equation*}
        X = \left\{\frac{1}{2^j}ight\}^\infty_{j=1} \bigcup \left\{0ight\}.
      \end{equation*}
      Then let $X'$ be the union of $X$ and smaller versions of $X$ in between each $1/2^n$, i.e.:
      \begin{equation*}
        X' = X \bigcup^\infty_{n=1}\left\{\frac{1}{2^{m+n+1}}+\frac{1}{2^{n+1}}ight\}^\infty_{m=1}.
      \end{equation*}
      This $X'$ is our compact set of real numbers whose limit points form a countable set.
    \end{claim}
    \begin{proof}
      The limit points of $X'$ form a countable set because by construction, the limit points are the points in $X$, which is countable since $j \in \mathbb{N}$.
      \par Now prove $X'$ is compact. Given any sequence $\{x_i\} \subset X'$, every $x_i$ is within the neighborhood of the closest number in the set $X$. This implies there is a subsequence $\{x_{ik}\}$ that converges to itself (if it is constant), or to a member of $X$. By the Sequential Compactness Theorem, this implies compactness since $\mathbb{R}$ is separable.
    \end{proof}

Exercise 2.13

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Exercise 2.13

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