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I am to prove that every bounded sequence in $\Bbb{R}$ has a convergent subsequence. I am stuck not knowing how and where to start.

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closed as off-topic by Carl Mummert, tired, Daniel W. Farlow, Dan, Nehorai Feb 19 at 21:03

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16 minutes. $ $ – Did Dec 23 '12 at 10:21
up vote 6 down vote accepted

This is the so called Bolzano-Weierstrass Theorem.

I will prove that any sequence of real numbers has a monotone subsequence and leave the rest to you

First some terminology: Let $\left(a_n\right)$ be a sequence of real numbers and $m\in \mathbb{N}$. We say that $m$ is a peak of the sequence $\left(a_n\right)$ if $$n> m\implies a_n<a_m$$

Now the proof: Let $\left(a_n\right)$ be a sequence of real numbers.

Suppose that $\left(a_n\right)$ has an infinite number of peaks $$k_0<k_1<k_2<...<k_n<...$$ and consider the subsequence $(a_{k_n})$. Then, $(a_{k_n})$ is strictly decreasing since $$k_n>k_m\implies a_{k_n}<a_{k_m}$$ and thus $(a_{k_n})$ is monotone.

Suppose that $\left(a_n\right)$ has an finite number of peaks and let $N\in \mathbb{N}$ be the last (greatest) peak. Then $k_{0}=N+1$ is not a peak and so \begin{equation}\exists k_1>k_0\in \mathbb{N}:a_{k_{1}}>a_{k_{0}}\end{equation} Having defined $k_n$ such as that $k_n>k_{n-1}>N$ then \begin{equation}\exists k_{n+1}>k_n\in \mathbb{N}:a_{k_{n+1}}>a_{k_{n}}\end{equation} The subsequence $(a_{k_n})$ is obviously increasing and so it is monotone

Now if $(a_n)$ is in addition bounded, so is $(a_{k_n})$ and applying the monotone convergence theorem yields....

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...yields that the sequence has finite limit – doniyor Dec 23 '12 at 10:13
@doniyor the subsequence – Nameless Dec 23 '12 at 10:14
yeah, the subsequence, thanks! – doniyor Dec 23 '12 at 10:15

Hint: The key result is the least-upper-bound property.

EDIT: I develop my argument. Let $(x_n)$ be a real bounded sequence and set $y_n= \sup \{ y_k : k \geq n \}$. Then $(y_n)$ is non-increasing sequence and you can show that it converges to $x=\inf \{ y_n : n \geq 0 \}$ from the definition of $\inf$. Finally, you can proved that $x$ is a limit point of $(x_n)$ and deduce that there exists a subsequence of $(x_n)$ converging to $x$.

In fact, least-upper-bound property, Bolzano-Weierstrass theorem and completness are very quite related in ordered fields.

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Since the sequence $(x_n)_n$ is bounded, every $x_n$ is in a bounded interval $I_0$. Split $I_0$ into two subintervals of half its length. At least one of them contains infinitely many terms of the sequence, call it $I_1$. Repeat. After $n$ steps, $I_n$ is an interval of length the length of $I_0$ divided by $2^n$, and $I_n\subset I_{n-1}$. Call $I_n=(a_n,b_n)$ and pick $x_{\varphi(n)}$ in $I_n$ with $\varphi(n)\gt\varphi(n-1)$ (crucially, this is always possible since $I_n$ contains infinitely many terms of the sequence). Then $(a_n)_n$ is nondecreasing, $(b_n)_n$ is nonincreasing, $b_n-a_n\to0$ and $a_n\leqslant x_{\varphi(n)}\leqslant b_n$, hence everything is set so that the subsequence $(x_{\varphi(n)})_n$ converges.

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