# About cardinality of cauchy sequences of rational numbers

Please could you see if I answered right to number (1) and (2), and give me a hint for (3). The question is:

Each one of the following sets has the cardinality of $\mathbb N$, $P(\mathbb N)$ and $P(P( \mathbb N))$. Classify them justifying your answer

1) The set of all Cauchy sequences $(a_n)$ of rational numbers.

2) The set of all Cauchy sequences $(a_n)$ of rational numbers with rational limit.

3) The set of all Cauchy sequences $(a_n)$ of rational numbers for which there exists a polinomial $p\in\mathbb Q[X]$ such that, for all $n\in\mathbb N$, $a_{n+1}=p(a_n)$.

For (1) Let $C$ be the set of all Cauchy sequences $(a_n)$ of rational numbers, and $A$ be the set of constant sequence of values $0$ ou $1$.

I mean, if $(x_n)\in A$, then $(x_n)$ is the constant sequence $0$ ou $1$. As every constant sequence is convergent, and every convergente sequence is a Cauchy sequence, $A\subset C$, proper subset, and we have $|A|<|C|$.

But $A=\{0,1\}^{\mathbb N}$, then $|A|=|\{0,1\}^{\mathbb N}|=2^{\mathbb N}=|P(N)|$, then $|P(N)|<|C|$.

For (2), let $B$ the set of all Cauchy sequences $(a_n)$ of rational numbers with rational limit, then $A\subset B\subset C$, where $A$ is a proper subset of $B$, then as in (1)

$|A|<|B|$, thus $|P(N)|<|B|$.

For (3), as $a_{n+1}=p(a_n)$ if the limit is $L$, then by continuity we have $L=p(L)$ and i do not know how to continue.

Thank you

EDITED

Number (1) is ok, it has $P(\mathbb N)|$, for number (3) I did this

Len $N$ be the set of Cauchy sequences as in (3), then if $x\in N$, then there exists $x_0\in x$ and a polynomial $p\in \mathbb Q[X]$ such that $x_n=p^n(x_0)$, for $n\in\mathbb N$.

Then, fixed $(x_0,p)\in x\times\mathbb Q[X]$ I define the function:

$f_{x_0,p}:\mathbb N\to \mathbb Q$, as $f_{x_0,p}(n)=p^{n}(x_0)$, i.e. $f_{x_0,p}(n)=x_n$. So I get another map

$$x\times \mathbb Q[X]\to N,(x_0,p)\mapsto x_n$$ wich is a bijection. There for $|N|=|x\times \mathbb Q[X]|=|\mathbb Q[X]|=|\mathbb N|$.

So number (2) is $|P(P(\mathbb N))|$, but i don't see how to prove it formally.

• When you define $A$ to be the set of constant sequences with values $0$ or $1$, that means to me that it has two elements: the $0$-sequence and the $1$-sequence. But if you define $A=\{0,1\}^{\mathbb{N}}$, then it also contains sequences which are not Cauchy, like $(0,1,0,1,0,1,\dots)$. Maybe it is better to consider for every sequence $x\in\{0,1\}^{\mathbb{N}}$ the sequence $y(x)$, where the $n$th term of $y$ is given by $(-1)^{x(n)}\cdot \frac{1}{n}$. Also, what can you say about the number of polynomials with rational coefficients? Jan 29, 2017 at 4:33
• Just because $A\subsetneq C$ doesn't imply $|A|<|C|$. The most you can conclude from that is $|A|\le|C|$. You're making this mistake in both of your parts (1) and (2). Jan 29, 2017 at 4:44
• For (3), note that each of these sequences is uniquely given by the combination of $a_0\in\mathbb Q$ and $p\in\mathbb Q[X]$. How many such combinations are there? Jan 29, 2017 at 4:49
• @martin.koeberl Thank you for your answer. I got it. For (3) $|\mathbb Q[X]|=|\mathbb N|$ and using the advise of Henning Makholm, I get the map
– John
Jan 29, 2017 at 14:49

You're misunderstanding the meaning of $\{0,1\}^{\mathbb{N}}$. As you've defined it, $A$ is "the set of constant sequences of values 0 [or] 1", and it therefore includes only two elements: the sequence $(0,0,0,\ldots)$ and the sequence $(1,1,1,\ldots)$.

$\{0,1\}^{\mathbb{N}}$, on the other hand, contains a huge number of sequences - every combination of zeroes and ones, including (for example) $(0, 1, 0, 1, 0, 1, \ldots)$. The vast majority of these are not Cauchy, so $\{0,1\}^{\mathbb{N}} \nsubseteq C$.

$|C| = 2^{\mathbb{N}}$, however; for any infinite sequence $a_n$ of zeroes and ones, we can define a sequence $x_n = (-1)^{a_n}2^{-n}$, which is Cauchy with limit $0$. Each of these sequences is different, so the map taking $(a_n)$ to $(x_n)$ is an injection from $\{0,1\}^{\mathbb{N}}$ into $C$. And $C$ can't be bigger than $2^{\mathbb{N}}$; it's a set of sets of rationals, and there are only $|\mathbb{N}|$ rationals.

This example can also be used for (2), and something similar for (3); I'll leave it to you to figure out how.

• Actually, $C$ is not a set of sets of rationals, but a set of subsets of $\mathbb N\times\mathbb Q$ -- but the cardinality you end up with is right, Jan 29, 2017 at 4:46
• Thank my friends, I wrote my solution for (3), can you give a look? Thank you
– John
Jan 29, 2017 at 15:17
• @John Your answer for (3) is now correct, but your answer for (2) is not. The set of rational Cauchy sequences with rational limit is a subset of the set of rational Cauchy sequences in general - but you've given it a larger cardinality. This isn't possible. Look at the example I gave demonstrating that (1) has cardinality $|P(\mathbb{N})|$, and think about what it means for (2) - notice that the sequences I used all have rational limits, because they all limit to $0$. Jan 29, 2017 at 15:40
• @Reese So there is an injection from $\{0,1\}^{\mathbb N}$ into $B$, and by the same argument, it has also cardinality $|P(\mathbb N)|$. I thought that each one of the three set had to have one of the cardinalities $|\mathbb N|$, $|P(\mathbb N)|$ ou $|P(P(\mathbb N))|$. Thank you very much!!!!
– John
Jan 29, 2017 at 15:58