0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ All theorems are true, so any of them implies any of them $\endgroup$ – vrugtehagel Feb 10 '16 at 10:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Watson Feb 10 '16 at 10:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Watson Feb 10 '16 at 10:56
1
$\begingroup$

Theorem 1 is what I know as the Bolzano-Weierstrass theorem. But there are no "official" names for theorems in mathematics, so I would not be surprised to learn that some people call the closely related Theorem 2 the Bolzano-Weierstrass theorem.

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 $x_n>x_{n_k}$.

$\endgroup$
0
$\begingroup$

(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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.