I have a question about a part of this proof I have:

We have two sequences such that:

$\forall n : a_n\le a_{n+1}\le b_{n+1}$ $a_n$ is a monotone increasing sequence bounded above by $b_{n+1}.$

$\forall n : b_n\ge b_{n+1}> a_{n+1}$ $b_n$ is a monotone decreasing sequence bounded below by $a_{n+1}.$

We know that every monotone and bounded sequence converges so, we'll show that $b_n\to c_b$ and $a_n\to c_a$. So $\lim(c_b-b_n)<\frac \epsilon 2 : \lim(c_a-a_n)<\frac \epsilon 2 $

Now we're left with showing that the limit is the same for both sequences: $b_n\to c_a$

$\lim(b_n-c_b)=\lim(b_n-a_n+a_n-c_b)\le \color{blue}{|\lim(b_n-a_n)|}+|\lim(c_b-a_n)|\le \frac \epsilon 2+\frac \epsilon 2=\epsilon$

My question is why the part marked in blue is $|\lim(b_n-a_n)|\le\frac \epsilon 2$ ?

NOTE: I may be wrong about the name of this proof, we call it just Cantor theorem and it's related to BW theorem.

  • 1
    $\begingroup$ What if $a_n=-(1/n)-1$ and $b_n=1/n$? These sequences satisfy you hipothesis but do not converge to the same limit. $\endgroup$ – Luiz Cordeiro Jun 6 '14 at 15:59
  • $\begingroup$ @LuizCordeiro your $a_n$ is not smaller than $a_{n+1}$. $\endgroup$ – GinKin Jun 6 '14 at 16:05
  • $\begingroup$ $n<n+1\Rightarrow 1/n>1/(n+1)\Rightarrow -1/n<-1/(n+1)\Rightarrow a_n=(-1/n)-1<(-1/(n+1))-1=a_{n+1}$. $\endgroup$ – Luiz Cordeiro Jun 6 '14 at 16:10
  • $\begingroup$ @LuizCordeiro, oh right, my bad. So is this a counter example for BW theorem ? $\endgroup$ – GinKin Jun 6 '14 at 16:25
  • $\begingroup$ Not really! What I think is that there is some previous information about the sequences $(a_n)$ and $(b_n)$ that you forgot to add here, probably something like $\lim(b_n-a_n)=0$. $\endgroup$ – Luiz Cordeiro Jun 6 '14 at 16:29

The conclusion of Cantor's theorem is that the infinite intersection $\bigcap_{n=1}^\infty [a_n,b_n]$ is not empty (in fact, this intersection is the interval $[c_a,c_b]$). As the example by @LuizCordeiro demonstrates, it is not necessary that $c_a = c_b$. One of the proofs of the Bolzano-Weierstraß theorem constructs a convergent subsequence of a given bounded sequence $(x_k)_{k=1}^\infty \subset [a,b]$ by successively choosing an interval $[a_n,b_n]$ containing infinitely many terms of $(x_k)_{k=1}^\infty$ so that $[a_n,b_n]$ is one of the two halves of the previous interval $[a_{n-1},b_{n-1}]$. This construction explicitly ensures that $b_n-a_n = 2^{-n}(b-a) \to 0$ as $n \to \infty$, so that we can later conclude that $c_a = c_b$ is the limit of the constructed subsequence. Below is a proof of Cantor's theorem.

The sequence $(a_n)_{n=1}^\infty$ is monotonically increasing and bounded from above by $b_1$, as $a_n \leq b_n \leq b_1$. Hence, $(a_n)_{n=1}^\infty$ converges to a finite limit $c_a \in \mathbb{R}$. Recall that $c_a:=\lim_{n\to\infty}a_n$ is in fact the least upper bound $\sup_n a_n$ of the set $\{a_n\;|\;n\geq1\}\subset \mathbb{R}$. Indeed, if we assume that $a_n>c_a$ for some $n$, then putting $\varepsilon=\frac12(a_n−c_a)>0$, we observe that for all $m \geq n$, we would have $a_m \geq a_n > c_a+ \varepsilon$, a contradiction. Also, if we assume that some $s<c_a$ is an upper bound for $(a_n)_{n=1}^\infty$, then putting $\varepsilon=\frac12(c_a−s)>0$, we obtain $a_n\leq s < c_a−\varepsilon$ for all $n$; again a contradiction. Similarly, we have $c_b := \lim_{n\to\infty}b_n = \inf_n b_n$, the greatest lower bound of the set $\{b_n\;|\;n\geq 1\}$. Now, observe that $a_m \leq b_n$ for arbitrary $m$ and $n$, not only when $m=n$. Indeed, for $p = \max\{m,n\}$ we have $a_m \leq a_p \leq b_p \leq b_n$. Fix an arbitrary index $n\geq1$ and consider the inequalities $a_m \leq b_n$ for all $m \geq 1$. They imply that $b_n$ is an upper bound for $(a_m)_{m=1}^\infty$. Hence, $c_a = \sup_m a_m \leq b_n$. Since $n$ was chosen arbitrarily, the last inequality holds for all $n \geq 1$. Thus $c_a$ is a lower bound for $(b_n)_{n=1}^\infty$, and hence, $c_a \leq \inf_n b_n = c_b$. Thus, the interval $[c_a,c_b]$ is not empty. Since $a_n \leq c_a \leq c_b \leq b_n$, we have $[c_a,c_b] \subset [a_n,b_n]$ for all $n$, or in other words, $[c_a,c_b] \subset \bigcap_{n=1}^\infty [a_n,b_n]$. Therefore, the latter intersection is non-empty, Q.E.D. In fact, we have also the inclusion $\bigcap_{n=1}^\infty [a_n,b_n] \subset [c_a,c_b]$, as every element of the intersection on the left is an upper bound for $(a_n)_{n=1}^\infty$ and simultaneously a lower bound for $(b_n)_{n=1}^\infty$, and hence $c_a \leq x \leq c_b$. Thus, $\bigcap_{n=1}^\infty [a_n,b_n] = [c_a,c_b].$

  • $\begingroup$ All right, but I still need the full and rigorous proof. I guess the proof I wrote above isn't good since I can't justify the part in question. $\endgroup$ – GinKin Jun 6 '14 at 17:11
  • $\begingroup$ Without further information about the sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$, nobody can justify the inequality $|a_n-b_n|<\varepsilon/2$ for all sufficiently large $n$, if that is what you mean. Your argument simply attempts to prove something that is not necessarily true. I will give a proof of the fact that $c_a \leq c_b$ (which is the actual conlusion of Cantor's theorem) in another comment (or possibly a couple of comments) below. $\endgroup$ – ivanpenev Jun 6 '14 at 17:38
  • $\begingroup$ Thanks. I think it would be easier for you to add this to your answer though. $\endgroup$ – GinKin Jun 6 '14 at 17:59
  • $\begingroup$ I added the proof in my answer, as you requested. I hope it is readable enough. $\endgroup$ – ivanpenev Jun 6 '14 at 18:38

Better to do this

$$|c_b-c_a| \leq |c_b-b_n|+|b_n-a_n|+ |a_n-c_a| \leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}$$

where for any given $\epsilon$ we chose $n$ large enough.

  • $\begingroup$ Still, why is $|\lim(b_n-a_n)|\le\frac \epsilon 3$ ? $\endgroup$ – GinKin Jun 6 '14 at 16:28

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.