On the proof of sequentially compact subset of $\mathbb R$ is compact

Question

I don't understand last steps of proving the following theorem:

My questions are:

1- Why the statements "cover index $(x_{n_k}) \le$ cover index $(x_0)$ for each index $k \ge K$" or/and "cover index $(x_{n_k}) \ge n_k \ge k$ for all indices k", contradicts with the statement "cover index $(x_n) > n$ for every index n"?

2- In the text before the Figure 2.6. it supposes in $I_n$, n is finite and consider some J. But the 3rd line after Eq.2.28, ignore that fact and by assumption n can be infinite as well. Wrong assumption? On the other hand, If n can be infinite also when defined J (before the Figure 2.6.), cover index $(x_0)$ (which appears as upper bound thereafter) can also be infinite, makes the statements "cover index $(x_{n_k}) \le$ cover index $(x_0)$ for each index $k \ge K$" or/and "cover index $(x_{n_k}) \ge n_k \ge k$ for all indices k" nonsense. Wrong assumption again?

Thank you.

EDIT - I shrink the two above question to one and make it clearer (and $CI$ stands for cover index):

For every $n\in \mathbb N$, there is (at least) one point (name it $x_n$) in $S$ such that $CI(x_n)>n$. The sequence ${\{x_n}\}$ has a sub sequence ${\{x_{n_k}}\}$ such that limit of that sub sequence belongs to $S$; by limit we mean when $n_k\rightarrow \infty$, so in fact $x_0$ is a 'name' for $x_{\infty}$; since $CI(x_n)>n$ so $[CI(x_{\infty})=]\ CI(x_0)>\infty$. Thus $\infty=?CI(x_0)> CI(x_{n_k})\ge x_{n_k}\ge k$ holds no matter how large $k$ becomes. No contradiction, contrary to the statement of the book.

Answer 1

Note: It seems the proof is correct, but a somewhat misleading notation may cause confusion.

The author uses the variable $n$ twice in different context. The first time $n$ is used as bounded variable in the expression

\begin{align*} S\subseteq \bigcup_{n=1}^NI_n \end{align*}

In the paragraphs next to Figure 2.6 the variable $n$ is used in a different context. Maybe the things become easier comprehensible, when we analyse this text by using another variable name $m$ instead of $n$.

If there is no natural number $N$ such that (2.26) holds, then for each natural number $m$, there is a point in $S$ whose cover index is greater than $m$; choose such a point and label it $x_m$. Thus, $\{x_m\}$ is a sequence in $S$ such that

\begin{align*} \text{cover index}(x_m)>m\qquad\qquad \text{for every index }m.\qquad\qquad (2.28) \end{align*}

Here we assume indirectly that $S$ is not compact and therefore no representation like (2.25) is possible. If no finite cover exists, we find to each natural number $m$ and the corresponding subcover $S_m := \bigcup_{j=1}^mI_j$ a point $x_m$ which is not covered by $S_m$. This way we obtain a sequence $\{x_m\}$ in $S$ explicitly constructed so that for each element $x_m$ its cover index is greater than $m$.

But, by assumption the set $S$ is sequentially compact. Thus, there is a subsequence $\{x_{m_k}\}$ that converges to a point $x_0$ that also belongs to $S$. As noted above, we can choose an open interval $J$ centered at $x_0$ such that (2.27) holds. However, $x_0$ is the limit of the sequence $\{x_{m_k}\}$, so there is an index $K$ such that \begin{align*} x_{m_k} \text{ belongs to }J\text{ for each index }k\geq K. \end{align*}

We consider the limit point \begin{align*} x_0:=\lim_{k\rightarrow \infty} x_{m_k} \end{align*} of the convergent subsequence $\{x_{m_k}\}_{k\geq 1}$. The notation $x_0$ is as suitable as $x_\infty$. We just need a symbol to uniquely identify the limit point of the subsequence.

This limit point $x_0$ is a point in $S$ and is therefore covered by $\bigcup_{j=1}^{M_0}I_j$ with $M_0$ being the cover index of $x_0$. We consider the interval $I_{M_0}$ and construct according to Figure 2.6 an interval $J_{M_0}$ centered around $x_0$ with

\begin{align*} x_0\in J_{M_0}\subseteq I_{M_0} \end{align*}

The clou: Since $x_0$ is the limit point of the subsequence $\{x_{m_k}\}$ all but finitely elements are in the Interval $J_{M_0}$. This implies that for all but finitely elements $x_{m_k}$ of this convergent subsequence their cover index is less than or equal $M_0$ and is therefore less than $m_k$ violating the construction principle of the sequence $\{x_{m}\}$.

Summary:

• (1) According to the indirect assumption that $S$ cannot by covered by a finite number of intervals, there is for each number $m$ an element $x_m\in S$ with a cover index $(x_m)>m$. On the other hand a convergent subsequence $\{x_{m_k}\}_{k\geq 1}$ of the sequence $\{x_m\}_{m\geq 1}$ has all but finitely elements within one specific interval $I_{M_0}$ implying that the cover index of these elements $x_{m_k}$ is less or equal $M_0$. So, for all but finitely elements of $\{x_{m_k}\}$ is the cover index$(x_{m_k})\leq M_0<m_k$ which is a contradiction.

• (2) The variable $n$ is used twice in different context.

Answer 2

Markus Scheuer in his answer has analyzed the given proof and verified that it is correct. It cannot be denied that this proof has shortcomings, and above all, the introduction of the covering index function makes it more complicated than necessary.

According to the standard definition a set $S\subset{\mathbb R}$ is compact if any given family ${\cal I}=(I_\alpha)_{\alpha\in I}$ of open intervals covering $S$ contains a finite subfamily $\bigl(I_{\alpha_k})_{1\leq k\leq N}$ covering $S$.

Assume now that $S\subset{\mathbb R}$ is sequentially compact and that we are given a covering of $S$ by a family ${\cal I}$ of open intervals $I_\alpha$. I claim that in any case the family ${\cal I}$ contains a countable subfamily ${\cal J}=\bigl(I_{\alpha_k})_{k\geq1}$ covering $S$.

Proof. For each $x\in S$ there are an $\alpha_x\in I$ and two rational numbers $p_x$, $q_x$ with $$x\in\ ]p_x,q_x[\ \subset I_{\alpha_x}\ .$$ The set ${\cal J}'$ of intervals $\ ]p_x,q_x[\ $ arriving in this way is countable. Select for each $J_k\in{\cal J}'$ an $x_k\in S$ with $J_k=\ ]p_{x_k},q_{x_k}[\ $, and put $\alpha_k:=\alpha_{x_k}$.$\quad\square$

Therefore we may now assume $S\subset \bigcup_{k\geq1} I_k$ with $I_k\in{\cal I}$ for all $k$. [This is the starting point in the quoted proof.] Put $$U_n:=\bigcup_{k=1}^n I_k\qquad(n\geq1)\ .$$ We shall show that there is an $N$ with $S\subset U_N$. Assume that this is not the case. Then for each $n\geq1$ we can find a point $x_n\in S\setminus U_n$. Since $S$ is sequentially compact there is a point $\xi\in S$ such that any neighborhood of $\xi$ contains points $x_n$ with arbitrarily large $n$. Now $\xi\in I_N$ for some $N$, and $I_N$ is a neighborhood of $\xi$. Therefore there is an $x_n\in I_N\subset U_N$ with $n>N$, contradicting the definition of $x_n$.$\quad\square$