# Short question about the proof of Cantor nested sets theorem

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.

• 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. – Luiz Cordeiro Jun 6 '14 at 15:59
• @LuizCordeiro your $a_n$ is not smaller than $a_{n+1}$. – GinKin Jun 6 '14 at 16:05
• $n<n+1\Rightarrow 1/n>1/(n+1)\Rightarrow -1/n<-1/(n+1)\Rightarrow a_n=(-1/n)-1<(-1/(n+1))-1=a_{n+1}$. – Luiz Cordeiro Jun 6 '14 at 16:10
• @LuizCordeiro, oh right, my bad. So is this a counter example for BW theorem ? – GinKin Jun 6 '14 at 16:25
• 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 $\lim(b_n-a_n)=0$. – 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, $\bigcap_{n=1}^\infty [a_n,b_n] = [c_a,c_b].$

• 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. – GinKin Jun 6 '14 at 17:11
• 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. – ivanpenev Jun 6 '14 at 17:38
• Thanks. I think it would be easier for you to add this to your answer though. – GinKin Jun 6 '14 at 17:59
• I added the proof in my answer, as you requested. I hope it is readable enough. – 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.

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