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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.

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    This is "essentially" (but not the same as) the [Bolzano-Weierstrass theorem](http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem). Do you know this theorem? What results are you allowed to use?2011-12-01
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    Have you considered proving $\mathbb{R} \cup \{ - \infty, + \infty\}$ is compact?2011-12-01
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    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.2011-12-01
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    Could you please expand that ideea, Isaac S.? I don't really understand that notion.2011-12-01
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    The question in the heading is ambiguous: It could mean "Is there any sequence that has at least one limit point?" or it could mean "Is it the case that any sequence, no matter which one, has at least one limit point." It appears that you mean the latter. Just changing "any" to "every" would make it completely unambiguous.2011-12-01
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    @MihaiBogdan So, if Bolzano-Weierstraß works for bounded sequences, maybe you can treat unbounded sequences separately.2011-12-01
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    Thanks Michael Hardy. Done that.2011-12-01
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    @Mihai, Michael suggested that you change "any" to "every" and not vice versa. =) [I had already edited the question to incorporate that change; it seems you have now reverted back to the original question.]2011-12-01
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    I feel so stupid. Sorry, I tried my best to express myself in English.2011-12-01

4 Answers 4

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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.

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    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.)2011-12-01
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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.). :-)

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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 $+\infty$.

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    Thank you Martin Sleziak, I understand this proof. Thanks, everyone.2011-12-01
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$\arctan : [-\infty,\infty] \rightarrow [-\pi/2,\pi/2]$. 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.