# Different versions of Bolzano Weierstrass Theorem and their relationships.

Which one is the Bolzano Weirerstrass Theorem?

Theorem 1. Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.

OR

Theorem 2. Every sequence of real numbers has a monotonic subsequence.

According to Wikipedia, Theorem 1 is the BW Theorem.

But according to this link, Theorem 2 is the BW Theorem.

I know there is another Theorem (let's call it Theorem 3) that states that every bounded monotonic sequence of real numbers has a finite limit. And it is straightforward that Theorem 2 and Theorem 3 implies Theorem 1.

My other question is: can we show that Theorem 1 implies Theorem 2? If yes, how can we prove it?

Helps are appreciated. Thank you!

• All theorems are true, so any of them implies any of them – vrugtehagel Feb 10 '16 at 10:18
• Theorem 1 is more general, in the sense that the theorem 2 requires the notion of "monotonicity", whereas Theorem 1 is true in every compact (metric) space. – Watson Feb 10 '16 at 10:21
• Actually, I shouldn't have put in brackets the word "metric" in my previous comment. For instance, the space $[0,1]^{\Bbb R}$ is a compact space, but not a sequentially compact space. However, a metrizable space is compact iff it is sequentially compact. – Watson Feb 10 '16 at 10:56

Here is a fairly easy way you can deduce Theorem 2 from Theorem 1. Let $(x_n)$ be a sequence in $\mathbb{R}$. If the sequence is unbounded above, then we can inductively define an increasing subsequence $(x_{n_k})$ by letting $n_0=0$ and letting $n_{k+1}$ be the least $n$ such that $x_n>x_{n_k}$. Similarly, if the sequence is unbounded below, we can get a decreasing subsequence. So let us assume the sequence is bounded. If there is some $a\in \mathbb{R}$ such that $x_n=a$ for infinitely many values of $n$, then we can just take the subsequence consisting of all such $x_n$. So let us also assume that the sequence does not repeat any value infinitely many times.
Now by Theorem 1, we may pass to a subsequence and assume $(x_n)$ converges to some $a\in\mathbb{R}$. Since $x_n=a$ for only finitely many values of $n$, either the set $S=\{n:x_n<a\}$ or the set $T=\{n:x_n>a\}$ must be infinite. Suppose $S$ is infinite; the other case is similar. Passing to the subsequence indexed by $S$, we may assume $x_n<a$ for all $n$. We now define a monotone increasing subsequence $(x_{n_k})$ by induction. First, let $n_0=0$. Given $n_k$, let $\epsilon=a-x_{n_k}$. Since $x_n$ converges to $a$, $|x_n-a|<\epsilon$ for all sufficiently large $n$, and since $x_n<a$, this means that $x_n>a-\epsilon=x_{n_k}$ for all sufficiently large $n$. We can thus define $n_{k+1}$ to be the least $n>n_k$ such that $x_n>x_{n_k}$.
(2)$\implies$(1) Let be $(x_{n_k})$ the monotonic subsequence. It will be also bounded. Take the $\sup$ of $(x_{n_k})$ if increasing and the $\inf$ if decreasing (completeness of $\Bbb R$ used here). You can check that the $\sup$/$\inf$ is the limit of $(x_{n_k})$.
(1)$\implies$(2) Suppose wlog that you start with a convergent sequence $(x_n)$. Also, you can suppose wlog that $\forall n\in\Bbb N: x_n>a =$ the limit. Start with $x_{n_1} = x_1$ and take $\epsilon_1 = x_1-a$. By the convergence, $\exists N(\epsilon_1)\in\Bbb N$ s.t. $|x_n-a|$ for $n\ge N(\epsilon_1)$. But this means that $l<x_n<x_1=a+\epsilon$ for $n\ge N(\epsilon_1)$. Any of this $x_n$ (for example, $x_{N(\epsilon_1)}$) as $x_{n_2}$. Repeat.