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Statement: Prove using the Bolzano-Weierstrass Theorem that any bounded increasing sequence in $\mathbb{R}$ converges.

Attempt: Let $\{a_n \}$ be a bounded increasing sequence. From the Bolzano-Weierstrass Theorem, there exists a subsequence $\{a_{n_k} \}$ which converges to some limit $A$. Then, $|a_n-A|\le|a_n-a_{n_k}|+|a_{n_k}-A|$. Let $\epsilon>0$ be given. Notice, $|a_{n_k}-A|<\frac{\epsilon}{2}$ for all $k\ge k_0$, for some $k_0\in\mathbb{Z}^+$, as $\{a_{n_k}\}$ converges to $A$. Also, since all convergent sequences are Cauchy, $|a_{n_k*}-a_{n_k}|<\frac{\epsilon}{2}$ for $k\ge k^*\ge N$, where $N\in\mathbb{Z}^+$. Since $\{a_{n_k}\}$ is a subsequence, there must exist an $n$ such that ${n_k*}\le n\le {n_k}$. Note that since $\{a_{n_k}\}$ is increasing, for all $k,k*\ge N$, $|a_{n}-a_{n_k}|\le|a_{n_k*}-a_{n_k}|<\frac{\epsilon}{2}$. Thus, $|a_n-A|<\epsilon$ for all $n\ge\max(n_{k_0},n_{k*})$. Thus, $\{a_n\}$ converges, as desired.

I saw a similar question during my search, but the answer used a different proof structure; curious if this is a valid proof.

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    $\begingroup$ In your proof, you may want to point out explicitly the part where you use monotonicity. For instance, at some point, the reader should see "By monotonicity of the sequence $(a_n)_n$, we have [...]". Indeed, you have to use monotonicity (the result is false without that assumption); so it should be immediately clear where you do use it. $\endgroup$ – Clement C. Jun 22 '18 at 22:00
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    $\begingroup$ You could make it simpler by using monotonicity, e.g., you can get rid of the absolute values in many inequalities, and you know that $0\leq A-a_n\leq A-a_{n_k}$ for all $n\geq n_k$. $\endgroup$ – Oskar Limka Jun 22 '18 at 22:35
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The problem here is that Bolzano theorem is proved by the monotonic limit theorem which states that a bounded monotonic sequence is convergent.

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    $\begingroup$ Not a problem, just shows that the two facts are equivalent. I believe it's entirely possible to construct real numbers starting from equivalent rational Cauchy sequences and proving BW before monotone sequence. $\endgroup$ – Oskar Limka Jun 22 '18 at 22:38
  • $\begingroup$ How do you prove Bolzano theorem? $\endgroup$ – hamam_Abdallah Jun 22 '18 at 23:15
  • $\begingroup$ In my class, we proved it using only that bounded sequences have a limit point, and then used the "lion chasing" method. $\endgroup$ – Saru Jun 22 '18 at 23:25
  • $\begingroup$ @Rasiel How do you prove that bounded sequences have a limit point. $\endgroup$ – hamam_Abdallah Jun 23 '18 at 0:06
  • $\begingroup$ @Salahamam_Fatima: all these facts are different expressions of completeness of real numbers. Given any one of them rest can can be proved. So it is more a matter of convenience (of textbook author) to choose anyone as the starting point for development of real analysis. $\endgroup$ – Paramanand Singh Jun 26 '18 at 2:48
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@Salahaman_Fatima here's the idea: first we need to define real numbers (starting from rationals). One way to do this, that uses monotonicity as a basis is via Dedekind cuts. Another, ultimately equivalent way, uses completion via Cauchy sequences. Each method has its pros and cons with respect to the other, I will describe Cauchy (to get to BW without monotone). Consider the the set \begin{equation} \mathfrak R:=\{\boldsymbol x=(x_n)_{n\in\mathbb N}\in\mathbb Q^{\mathbb N}:\boldsymbol x\text{ is Cauchy in }\mathbb Q\} \end{equation} and declare two elements $\boldsymbol x$ and $\boldsymbol y$ of $\mathfrak R$ to be equivalent if and only if their difference is a vanishing sequence. The ensuing quotient set is declared to be $\mathbb R$ where the Cauchy criterion can be shown to be equivalent to convergence. Necessity results from necessity in $\mathbb Q$, sufficiency requires a bit more work. Consider a Cauchy sequence of real numbers $(\xi_n)_{n\in\mathbb N}$, this defines a double sequence of rational numbers $x_{n,k}$ where $\xi_n=[(x_{n,k})_{k\in\mathbb N}]$ where the square bracket indicates the equivalence class with respect to the equivalence described above. Using the fact that $\xi_n$ is Cauchy for $n\in\mathbb n$ we can build a sequence $y_n:=x_{n,\hat k(n)}$ of rational numbers that is Cauchy, i.e., in $\mathfrak R$, hence the equivalence class $\eta:=[(y_n)_{n\in\mathbb N}]$ is in $\mathbb R$ and $\xi_n\to\eta$ can be now proved and established the Cauchy Criterion. To prove BW you work by bisection: a bounded sequence $\xi_n$, $n\in\mathbb N$, is inside a bounded interval $[a_0,b_0]$, pick $c_0:=(b_0+a_0)/2$ and define $[a_1,b_1]:=[a_0,c_0]$ if the latter has infinitely many terms (with repetition) of the sequence or $[a_1,b_1]:=[c_0,b_0]$ in the other case (because the sequence is infinite one of the two must occur because strictly speaking a sequence is a function from $\mathbb N$ via which the counterimages of the two intervals must cover all of $\mathbb N$ so one is infinite). By induction, you can now show that repeating iteratively this procedure provides you with a Cauchy sequence in $\mathbb R$ and this must be convergent by CC.

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  • $\begingroup$ But how do you prove that $\Bbb R$ is complete. I think you will need BW. $\endgroup$ – hamam_Abdallah Jun 26 '18 at 13:47
  • $\begingroup$ @Salahamam_Fatima please read my main post, it shows first that $\mathbb R$, defined as the quotient of $\mathfrak R$ by the vanishing difference equivalence, is complete (Cauchy criterion), without resorting to BW. Second, it shows how to prove BW by bisection. This is not in contradiction with what I think you are trying to say, i.e., that BW implies the Cauchy criterion. $\endgroup$ – Oskar Limka Jul 16 '18 at 8:48

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