# Does every sequence have at least one limit point?

Let $x_n$ be a sequence of real numbers.

Definition: $x \in \mathbb R \cup \{-\infty,\infty\}$ is a limit point of a sequence $x_n$ if there is a subsequence $x_{n_k}$ of our sequence such that $x_{n_k} \to x$.

If $A= \{x \in \mathbb R \cup \{-\infty,\infty\} \mid x \text { is a limit point of }x_n\}$, prove that $A$ is not empty.

• This is "essentially" (but not the same as) the Bolzano-Weierstrass theorem. Do you know this theorem? What results are you allowed to use? Dec 1, 2011 at 19:08
• Have you considered proving $\mathbb{R} \cup \{ - \infty, + \infty\}$ is compact? Dec 1, 2011 at 19:10
• Isn't Bolzano-Weierstrass theorem for bounded sequences? Well, I am allowed to use pretty much everything I know. I am only starting to learn topology. Dec 1, 2011 at 19:12
• Could you please expand that ideea, Isaac S.? I don't really understand that notion. Dec 1, 2011 at 19:16
• @MihaiBogdan So, if Bolzano-Weierstraß works for bounded sequences, maybe you can treat unbounded sequences separately. Dec 1, 2011 at 19:18

A proof sketch assuming Bolzano-Weierstrass.

1. Show that $x_n$ is bounded above if and only if $\infty$ is not a limit point of $x_n$.

2. Reversing the above argument, show that $x_n$ is bounded from below if and only if $- \infty$ is not a limit point.

3. From (1.) and (2.), conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ is bounded. Further, using Bolzano-Weierstrass, conclude that if neither $\infty$ nor $-\infty$ is a limit point of $x_n$, then $x_n$ has a subsequence converging to some limit $x \in \mathbb R$.

To summarize the argument, the key idea is that the sequence is

• either unbounded, in which case one of $+\infty$ and $-\infty$ is a limit point;
• or it's bounded, in which case Bolzano-Weierstrass shows the existence of a limit point in $\mathbb R$.

Note: Martin's answer explains the idea behind (1.). :-)

If the sequence is bounded, then you can invoke Bolzano-Weierstrass theorem.

So we can assume it is unbounded. Let us assume it is not bounded from above. (The symmetric case is similar.)

Thus for every $C$ there exists $n$ such that $x_n>C$.

Start with $C_1=1$ and choose $n_1$ such that $x_{n_1}>C_1=1$.

Now choose $C_2=2+\max\{x_j\; : \; j\le n_1\}$. Obviously $C_2\ge 2$ and you can choose $n_2$ such that $n_2>n_1$ and $x_{n_2}>C_2$.

You continue by induction. In the $k$-th step you choose $C_k=k+\max\{x_j \; : \; j\le n_k\}$. This means that $C_k\ge k$ and you there exists $n_{k+1}>n_k$ with $x_{n_k}>C_k$.

In this way you obtain a subsequence $n_k$ such that $x_{n_k} \ge k$. This implies that this subsequence converges to $+\infty$.

• Thank you Martin Sleziak, I understand this proof. Thanks, everyone. Dec 1, 2011 at 19:28

$\arctan : [-\infty,\infty] \rightarrow [-\pi/2,\pi/2]$. The sequence $\{x_n\}_{n=1}^\infty \subseteq \mathbb{R}$ has a limit point in $[-\infty,\infty]$ if and only if the sequence $\{\arctan x_n\}_{n=1}^\infty \subseteq [-\pi/2,\pi/2]$ has a limit point in $[-\pi/2,\pi/2]$. Bolzano-Weierstrass takes care of that.

Define $y_n = \sup \{ x_n, x_{n+1}, x_{n+2}, \cdots \} .$ Then $y_n$ is a monotone decreasing sequence and hence $Y = \displaystyle \lim_{n\to\infty} y_n$ exists (possibly $-\infty$).

The claim is that $Y \in A .$ If $Y$ is finite, then argue as follow: For any $\epsilon > 0$, by the definition of supremum for each $n\in \mathbb{N}$ we can find $n_k \geq n$ such that $|y_n - x_{n_k} | < \frac{\epsilon}{2}.$ By definition of limit, for some $n_0 \in\mathbb{N}$ we have $|Y - y_n| < \frac{\epsilon}{2}$ for all $n > n_0.$ Thus for all $n> n_0$, $$|Y - x_{n_k}| \leq |Y-y_n| + |y_n - x_{n_k}| < \epsilon.$$ Thus, $x_{n_k} \to Y.$

I leave the case where $Y= -\infty$ to you.

• Note that it is possible that all $y_n$'s are equal to $+\infty$. (E.g. the sequence $x_n=n$.) So perhaps you should deal with the case $Y=\infty$ too. (Or at least you should write that this case is also left to the reader.) Dec 1, 2011 at 20:05