3.3 Closed sets

This begs for the question: If we come across with a set $S$ of real numbers, can we decide whether there exists some sequence $\overline{x}={\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that $LIM(\overline{x})=S$. The following proposition gives an important condition for being limitset.

Proof:  We already had versions of this proposition: in the example when we showed that $Q$ is not a limitset and in the proof of the fact that Limsup is always a limitpoint. We use exactly the same idea. Since ${\{y_k\}}_{k=1}^{\mathrm{\infty }}$ are limitpoints, we have a sequence such that So, by the Sum Rule ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=y$. That is it. $\mathrm{\square }$

By the previous proposition any limitset $S$ must have the following property: Now matter how we choose a convergent sequence $\overline{x}={\{x_n\}}_{n=1}^{\mathrm{\infty }}$ from $S$ the limit of $\overline{x}$ is in the set $S$. This was of course true for the three examples in Kind Reminder 3.3.1. Also, it does not hold for the set of rational numbers and the set of irrational numbers as well (yes, you can approximate irrational numbers by rationals AND any rational number $q$ by the sequence of irrationals :${\mathrm{\{}\frac{\mathrm{1}}{n}\mathit{}\sqrt{\mathrm{2}}\mathrm{+}q\mathrm{\}}}_{n\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}\mathrm{.}$ The following definition is extremely important.

The following two propositions help you to find even more complicated closed sets.

Proof:  We only need to show that if ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ is a convergent sequence tending to $a$ such that for any $n\ge 1$ $a_n\in {\cap }_{k=1}^{\mathrm{\infty }}{F}_k$, then $a\in {\cap }_{k=1}^{\mathrm{\infty }}{F}_k$ as well. This is almost trivial. For any $n\ge 1$ and $k\ge 1$ $a_n\in F_k$, so $a\in F_k$ since $F_k$ is closed. The fact that $a$ is an element of all the $F_k$’s means exactly that $a\in {\cap }_{k=1}^{\mathrm{\infty }}{F}_k$. $\mathrm{\square }$

Proof:  Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be a convergent sequence tending to $a$ such that $a_n\in F\cup G$. We need to show that $a\in F\cup G$. Clearly, either there is a subsequence ${\{a_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ so that all the $a_{n_k}$’s are in $F$ or there is a subsequence ${\{a_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ so that all the $a_{n_k}$’s are in $G$. The limit of this subsequences is $a$ of course. So $a$ is either in $F$ or in $G$, that is $a\in F\cup G$. $\mathrm{\square }$

Proof:  This seems like a new proposition. It is not. It is just Proposition 3.3.1 in a new cloth. We proved that the limit of a convergence sequence of limitpoints of $\overline{x}$ is again a limitpoint. This means that the set of limitpoints of $\overline{x}$ is a closed set, not less, not more. $\mathrm{\square }$

Proof:  The proof is given by an algorithm. We take first the ten $\frac{1}{10}$-intervals, then the hundred $\frac{1}{100}$-intervals and so on. So, we will have the system of intervals $I_1,I_2,I_3,\mathrm{\dots }$. Our algorithm works the following way. Let us pick the first interval $I_{k_1}$ that contains an element of $F$ and let this element be $x_1$. Then go further, pick the next interval $I_{k_2}$ that contains an element of $F$ and let this element be $x_2$. Continuing the process we will obtain a sequence $\overline{x}={\{x_n\}}_{n=1}^{\mathrm{\infty }}$. Then $\text{LIM}(\overline{x})=F$. Why? We need to show two things.

1. 1.

   All elements of $F$ are limitpoints of $\overline{x}$.

2. 2.

   If $x\notin F$ then $x$ is not a limitpoint of $F$.

The second statement is trivial. Indeed, if ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ a convergent subsequence tending to $x$ then all the $x_{n_k}^{\prime }s$ are in our closed set $F$, so by the very definition of closedness $x$ must be in $F$ as well. The first statement is a bit more trickier, but not much. Let $x\in F$. Then $x$ is in the $\frac{1}{10}$ -interval $I_{m_1}$, in the $\frac{1}{100}$ -interval $I_{m_2}$, in the $\frac{1}{1000}$ and so on. When our algorithm run we cannot avoid this intervals. We must pick a point from each of them. So we will have a subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ such that $x_{n_k}\in I_{m_k}$. That is $|x_{n_k}-x|\le \frac{1}{{10}^k}$, hence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ converges to $x$. $\mathrm{\square }$

We have seen many kind of closed sets. Simple ones, complicated ones, finite ones and infinite ones. There is a very simple way to construct closed sets from arbitrary sets. As we have seen above for any closed set $F$, there exists a sequence of elements of $F$ such that the set of limitpoints of the sequence is exactly $F$. So, it we start with this sequence, we can reconstruct $F$. The following theorem is a variation of this theme.

Proof:  Let $s\in S$. Then $s,s,s,s,s,s,s,\mathrm{\dots }$ is a perfectly legitimate convergence sequence of elements of $S$ and its limitpoint is $s$ of course. So, $S\subset CL(S)$. Now, we prove that $CL(S)$ is closed. Let ${\{l_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence from $CL(S)$ tending to $l$. We need to prove that $l\in CL(S)$. Since, $l_n$ is in $CL(S)$, there exists $s_n\in S$ such that . By the Sum Rule, the sequence ${\{s_n\}}_{n=1}^{\mathrm{\infty }}$ is convergent and its limit is $l$. Hence, $l$ is element of $CL(S)$. Finally, suppose that $S\subset C$, where $C$ is some closed set. We need to prove that if $l\in CL(S)$ then $l\in C$. What does it mean that $l\in CL(S)$? It means, that there exists a convergent sequence ${\{s_n\}}_{n=1}^{\mathrm{\infty }}$ of elements of $S$ tending to $l$. This convergent sequence is in $C$, $C$ is closed, hence $l$ must be in $C$. $\mathrm{\square }$