2.2 Cauchy-sequences and the Bolzano-Weierstrass
Theorem

Proof:  By Proposition 4.5.2, if ${\{q_n\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy-sequence of RATIONAL numbers, then there exists a real number $x=x_0.x_1{x}_2\mathrm{\dots }$ such that $(q_n-{(x)}_n)\to 0$. Since ${(x)}_n\to x$, by the Sum Rule, $q_n\to x$ as well. So, at least it is clear that any Cauchy-sequence of rational numbers is convergent to a real number, its “limitvalue” (from now on we will not use the word limitvalue, since it is just a complicated name for the limit). Now, we need the following lemma.

Proof:  We need to prove that for any $\epsilon >0$ there exists $N>0$ such that if $n,m\ge N$, then $|(y_n-z_n)-(y_m-z_m)|\le \epsilon$. Since ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy-sequence and $z_n\to 0$, we have an $N>0$ so that

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  If $n,m\ge N$, then $|y_n-y_m|\le \frac{\epsilon }{2}.$

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  If $n\ge N$, then

Therefore by the Triangle Inequality, if $n,m\ge N$, then

As we have seen earlier, for any real number $x$ there is a convergent sequence of rationals tending to $x$. So, if ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy-sequence of real numbers then we can choose rational numbers ${\{q_n\}}_{n=1}^{\mathrm{\infty }}$ such that , that is $(x_n-q_n)\to 0$ by the Sandwich Rule. Therefore, by the Lemma ${\{q_n\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy-sequence of rational numbers tending to some number $x$. Since $x_n=q_n+(x_n-q_n)$, by the Sum Rule ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is convergent and its limit is $x$. $\mathrm{\square }$

If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence of real numbers, then for any sequence , ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ is called a subsequence of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$.

Solution: How does negation work? It is not true that some statement hold for all $\epsilon >0$. So, there exists an $\epsilon >0$ such that the statement doesn’t hold. This latter statement is that there exists some $N>0$ such that for any $n\ge N$ $|x_n-x|\le \epsilon$. The negation of this statement is that for ALL $N$ there exists some $n\ge N$, such that $|x_n-x|>\epsilon$. This means that there exists a subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ such that for any $k\ge 1$ $|x_{n_k}-x|>\epsilon$. That is, the negation of the statement that $x_n\to x$ is the following statement:

There exists some $\epsilon >0$ and a subsequence ${\{x_{n_k}\}}_{k=1}^{\mathrm{\infty }}$ such that for any $k\ge 1$, $|x_{n_k}-x|>\epsilon$.

Proof:  Let $M$ be an integer such that $|x_n|\le M$ for any $n\ge 1$. What can be the integer part of the numbers $x_n$? Well, something in between $-M$ and $M$. If you put infinitely many numbers into finitely many pigeon-holes then at least one of the pigeon holes will contain infinitely many numbers. Let $\overline{z_1}=x_{m_1^1},x_{m_2^1},x_{m_3^1}\mathrm{\dots }$ be a subsequence such that all of the $m_i^1$’s are in the form of $y.******$. (You will see immediately, why we use this complicated looking indices.) Then, infinitely many of the real numbers $x_{m_1^1},x_{m_2^1},x_{m_3^1}\mathrm{\dots }$ must have the same first digit $y_1$ (after the dot). Indeed, there are $10$ choices for the first digits and we have infinitely many numbers. Let $\overline{z_2}=x_{m_1^2},x_{m_2^2}\mathrm{\dots }$ be a subsequence of $x_{m_1^1},x_{m_2^1},x_{m_3^1}\mathrm{\dots }$ such that the first digit of the $x_{m_i^2}$’s are the same. So, $x_{m_1^1},x_{m_2^2},x_{m_3^2}\mathrm{\dots }$ start with the same integer $y$, $x_{m_1^2},x_{m_2^2}\mathrm{\dots }$ starts with $y$ and then the first digit is $y_1$. Then we continue this process and construct inductively for each $k\ge 1$ a subsequence $\overline{z_k}=x_{m_1^k},x_{m_2^k},x_{m_3^k}\mathrm{\dots }$, such that

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  $\overline{z_k}$ is a subsequence of $\overline{z_{k-1}}$.

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  For any $i\ge 1$, $x_{m_i^k}$ starts with $y.y_1,y_2\mathrm{\dots }{y}_k$.

Then we make our precious convergent subsequence by picking the first guy from each sequence $\overline{z_k}$, so $x_{a_k}=x_{m_1^k}.$ This sequence is a Cauchy-sequence, since if $n\ge N$, then $|x_{a_n}-x_{a_m}|\le \frac{1}{{10}^{-k}}.$ Hence this sequence is convergent. (we can see immediately, that ${\lim }_{k\mathrm{\to }\mathrm{\infty }}\mathit{}{a}_k\mathrm{=}y\mathrm{.}{y}_{\mathrm{0}}\mathit{}{y}_{\mathrm{1}}\mathit{}{y}_{\mathrm{2}}\mathit{}\mathrm{\dots }$). $\mathrm{\square }$

Instead of jumping to the next theorem, let us try to digest the proof of the Bolzano-Weierstrass Theorem (relax, it is one the hardest theorem in Math113, you are not supposed to understand it for the first sight, but if you really understand it, you will have a rather good picture of the notion of convergence). To use the word digits is kind of awkward, so we will use a more elegant language. This language will:

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  Help you to get rid of the “start with” and the “digits”, which are soooo high school.

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  Help you to visualize the proof of the Bolzano-Weierstrass Theorem.

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  Help you to prove trickier things.

How does the Bolzano-Weierstrass Proof work? By boundedness, we realize that there is an integer interval containing an infinite subsequence of our bounded sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$. We forget all other elements of the sequence. The first element of this subsequence will be $x_{a_1}$. Now, we realize that there is a ${10}^{-1}$ interval inside our integer interval that contains infinitely many elements of our first subsequence. This will be our second subsequence. We forget all other elements and decide the the first element of the second subsequence will be $x_{a_2}$. And so on. Then $x_{a_1},x_{a_2},x_{a_3}\mathrm{\dots }$ will be a Cauchy-sequence since $x_{a_n},x_{a_{n+1}},\mathrm{\dots }$ are all in the same ${10}^{-n}$-interval.