3.1 Limitpoints

Of course, if $\overline{x}={\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is convergent then it has one single limit point.

Proof:  Recall that $sup(\overline{x})$ is the the Least Upper Bound of $\overline{x}$, so if $z>\overline{x}$, then there is no $x_n$ such that $x_n>sup(\overline{x})+\frac{z-sup(\overline{x})}{2}.$ Therefore $z$ cannot be a limitpoint of $\overline{x}$. Hence, any limitpoint of $\overline{x}$ is less or equal than $sup(\overline{x})$. Similarly, any limitpoint of $\overline{x}$ is greater or equal than $sup(\overline{x})$. $\mathrm{\square }$

Proof:  By Question 2.4.2, a sequence of positive real numbers does not tend to infinity IF AND ONLY IF it has a bounded subsequence. However, if it has a bounded subsequence, then by the Bolzano-Weierstrass Theorem it must have a limitpoint. Now suppose that $\overline{x}$ is convergent to $y$. Then any subsequence of $\overline{x}$ is convergent to $y$ as well, hence $\overline{x}$ has only one limitpoint. Now suppose that $\overline{x}$ has one single limitpoint $y$, we need to show that $\overline{x}$ is converging to $y$. If $\overline{x}$ is not converging to $y$, then by Question 2.2.1, there exists $\epsilon >0$ and a subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ such that for any $k\ge 1$, $|x_{n_k}-y|\ge \epsilon$. Then, by the Bolzano-Weierstrass Theorem, we have a convergent subsequence of ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ tending to some $y^{\prime }$, so that $|y^{\prime }-y|\ge \epsilon$. This is in contradiction with the fact that $\overline{x}$ has only one limitpoint. $\mathrm{\square }$

Solution: Yes. Let $x_n=n$ if $n$ is even and $x_n=\frac{1}{n}$ if $n$ is odd. $\mathrm{\square }$

Observe that the sequence ${\{{(-1)}^n\}}_{n=1}^{\mathrm{\infty }}$ has two limit points: $0$ and $1$. Can you construct a sequence having exactly three limit points? (Yes, you can.)

Proof:  By the Sum Rule, a subsequence of ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ converges to $y$ if and only if ${\{x_{n_k}+y_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ converges to $y$. Hence the limitpoints of $\overline{x}$ and $\overline{x}+\overline{y}$ coincide. $\mathrm{\square }$

Solution: Let $L$ be the set consisting of the numbers ${\{\frac{1}{n}\}}_{n=1}^{\mathrm{\infty }}$ and $0$. This is an infinite set and it is the set of limitpoints of certain sequences. A sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ can be constructed as follows. Let $x_n=0$ if $n$ is odd. Let $x_n=\frac{1}{l}$ if $n$ can be divided by $\frac{1}{2^l}$ but $n$ cannot be divided by $\frac{1}{2^{l+1}}$. Then, the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ contains any number $\frac{1}{l}$ infinitely many times, hence we can form constant (!) subsequences converging to $\frac{1}{l}$. Clearly, $0$ is a limitpoint of a subsequence as well. If $y\notin L$, then there exists an $\epsilon >0$ such that the interval $[y-\epsilon ,+\epsilon ]$ does not contain a single element of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ so, it cannot be a limitpoint of a subsequence of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$. Note that there are many other examples, e.g. if you have an example you can add a sequence that converges to zero and you immediately obtain an other sequence. The number theoretical trick is cute, but any other sequence that contains all elements of the set $L$ infinitely many times would be good.

Solution: Yes, there are sequences like that. Remember, the set of rationals $\mathbb{Q}$ is countable. Countable means you can actually “count” or “enumerate” them, so let $q_1,q_2,q_3,\mathrm{\dots }$ be an enumeration of all the rationals. Now we repeat the trick of the previous question, we construct a sequence that contains all rational numbers infinitely many times. E.g: $x_n=q_l$ if $n$ can be divided by $\frac{1}{2^l}$ but $n$ cannot be divided by $\frac{1}{2^{l+1}}$. OK, it is clear that the limitpoints of this sequence contain all the rational numbers. However, ANY real number is a limitpoint of this strange sequence. Why? Let $r\in ℝ$ be a real number. Then we know that there exists a sequence of rational numbers ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ converging to $r$. Now we pick a subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ such that $x_{n_k}=a_k$. Then the subsequence will converge to $r$.

Solution: The answer is NO. As we have seen in the solution of the previous question, one can construct a sequence $\overline{x}$ such that all the rationals are limitpoints of $\overline{x}$, but…. No matter how clever is our construction, $\sqrt{2}$ (and as a matter of fact any other irrational number $r$) will also be a limitpoint of the sequence. So, suppose that $\text{LIM}(\overline{x})=\mathbb{Q}$. Let ${\{q_k\}}_{k=1}^{\mathrm{\infty }}$ be a sequence of rational numbers converging to the irrational number $r$. If the sequence $\overline{x}$ contains each of these $q_k$’s infinitely often we immediately see the contradiction. However, it is possible that not a single $q_k$ is an element of our sequence. It does not matter. The point is that all the rational numbers $q_k$ are limitpoints of $\overline{x}$. Thus, there exists $n_1>0$ such that . Then, there exists $n_2>n_1$ such that . And so on, there exists such that . By the Sum Rule, the subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ converges to the same guy as the sequence ${\{q_k\}}_{k=1}^{\mathrm{\infty }}$ does, and…. this guy is just our irrational number $r$.