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I have a problem with this exercise:

How many sequences of rational numbers converging to 1 are there?

I know that the number of all sequences of rational numbers is $\mathfrak{c}$. But here we count sequences converging to 1 only, so the total number is going to be less. But is it going to be $\mathfrak{c}$ still or maybe $\aleph _0$?

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11  
If $x_n\rightarrow0$, then $1\pm x_n\rightarrow 1$ for any choice of signs. – David Mitra Jan 9 at 14:02
    
So the number of the sequences converging to 1 is equal to number of the sequences converging to 0. But what now? – Bartłomiej Sługocki Jan 9 at 14:07
    
encode in $x_n$ a sequence of bits, so that the answer is $\ge$ than $2^{\aleph_0} = \aleph_1$. then considering a general sequence of rationnal numbers (not constrainted to converge to $1$) show that the answer is $\le$ than $2^{\aleph_0} = \aleph_1$ – user1952009 Jan 9 at 14:07
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@user1952009 $2^{\aleph_0}$ isn't necessarily $\aleph_1$ - that equality is the continuum hypothesis, which is independent of ZFC. ($2^{\aleph_0}$ does however equal $|\mathbb R|$) – Milo Brandt Jan 9 at 17:20
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How many infinite subsets of $\{1+\frac1n:n\in\Bbb N_+\}$ are there? – Akiva Weinberger Jan 10 at 22:05
up vote 4 down vote accepted

We have that non-repeating (injective) sequences of elements in $\{\,1+1/n:n\in\mathbb{N}\,\}$ form a continuum, and all of them have limit $1$, so our set is at least a continuum. Since also all rational sequences form a continuum, our set is also at most a continuum.

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The number of sequence of rational numbers converging to $1$ is not countable. Suppose you get all squences by $(a_n^{(1)})_{n\in \mathbb{N}}, (a_n^{(2)})_{n\in \mathbb{N}}, (a_n^{(3)})_{n\in \mathbb{N}}, \dotsc$ Define a sequence $(b_n)_{n\in \mathbb{N}}$ by $$b_{k}:=\begin{cases}1 & a_k^{(k)}\neq 1\\ 1+\frac{1}{n}& a_k^{(k)}= 1 \end{cases}$$ Then $\lim_{n\to\infty}b_n=1$ but $(b_n)_{n\in \mathbb{N}}\neq (a_n^{(i)})_{n\in \mathbb{N}} $ for all $i\in\mathbb{N}$.

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Is it possible to prove it's $2^{\aleph_0}$ without CH? – Kevin Jan 10 at 6:34
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@Kevin: Yes. At position $n$ in the sequence, you can have $1+\frac 1n$ or $1$. That is in bijection with the reals in $(0,1)$ expressed in binary. – Ross Millikan Jan 10 at 16:51

For each real number $x$ define $$a_n(x):= 1+ \frac{ \lfloor xn \rfloor}{n^2}$$

Show that this is a one-to-one function from $\mathbb R$ to the set of sequences converging to $1$.

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@DavidRicherby That means that the function is not ONTO... I never said anything about that ;) – N. S. Jan 10 at 19:04
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Oh. Duh. I hate the term "one-to-one" because, to me, it sounds much more like it means "bijective" than "injective" so I always make that mistake. – David Richerby Jan 10 at 21:14

I believe there is a continuum number of such sequences. For ease of writing, let's try to equivalently count the number of sequences converging to $0$. First note that there are countably many rational numbers in the interval $[-a,a]$. Now set for every sequence we will consider $x_i \in [\frac{-1}{i},\frac{1}{i}]$, so that each sequence of $x_i$'s will converge to $0$. The number of such sequences is equivalent to the number of functions from $\mathbb{N}$ to itself, which is provably uncountable.

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the question was indeed how to prove that the number of such sequences is the number of functions from $\mathbb{N}$ to itself ? – user1952009 Jan 9 at 17:43
    
You have demonstrated there is at least a continuum of such sequences. Can you provide an upper bound? – Jan Dvorak Jan 10 at 14:02
    
@JanDvorak: OP correctly said the number of all sequences of rationals is $\mathfrak c$. The ones converging to $0$ are a subset of these, so $\mathfrak c$ is an upper bound. – Ross Millikan Jan 10 at 16:47

$q \pm \frac{1}n$ is a sequence converging to $q$ for any sequence of $\pm$ signs.

That is a lower bound matching the stated upper bound. This solves the problem in the sense that the Schroeder Bernstein principle applies. Finding a specific 1-1 correspondence between rational convergent sequences and 0-1 sequences is more complicated.

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Isn't simply $\mathbb{N}^\mathbb{N}$ a continuum, whence also $\mathbb{Q}^\mathbb{N}$ is? – yo' Jan 10 at 21:53
    
The convergent sequences are a superset of a copy of 2^N from the +- signs, and a subset of a Q^N. But this is a Schroeder-Bernstein argument, not a direct bijection to 2^N. – zyx Jan 15 at 0:33

Suppose you have a sequence of rationals $(a_i)$ that converges strictly monotonically to $1$ from above, that is, for all $i$, $a_i > a_{i+1}$. Consider an arbitrary sequence of non-negative integers $(c_k)$. Construct a subsequence of $(a_i)$ by skipping the first $c_0$ elements of $(a_i)$ and set $b_0$ to the next element of $(a_i)$; then skip another $c_1$ elements of $(a_i)$ and set $b_1$ to the next element of $(a_i)$; then skip another $c_2$ elements of $(a_i)$, and so forth. This produces a sequence $(b_k)$ which is a subsequence of $(a_i)$.

Any two distinct sequences of integers $(c_k)$, $(d_k)$ will produce two distinct subsequences of $(a_i)$ via this procedure. For if $m$ is the least integer such that $c_m \neq d_m$, the first $m$ elements of the subsequences produced for $(c_k)$ and $(d_k)$ will be equal, but the next elements will be different, since we skip different numbers of elements to find the next element of each subsequence, and no two elements of $(a_i)$ are equal.

That is, following this procedure, for every sequence of non-negative integers there is a unique sequence of rationals that converges to $1$. Let $A$ be the set of all sequences of non-negative integers and $S$ be the set of all sequences of rationals converging to $1$; we have shown that $\left\vert{A}\right\vert \leq \left\vert{S}\right\vert$.

But of course every sequence of rationals converging to $1$ is a sequence of rationals, so if $B$ is the set of all sequences of rationals, $\left\vert{S}\right\vert \leq \left\vert{B}\right\vert$.

But we also know that $\left\vert{A}\right\vert = \left\vert{B}\right\vert = \mathfrak{c}$. That is, $\mathfrak{c} \leq \left\vert{S}\right\vert$ and $\left\vert{S}\right\vert \leq \mathfrak{c}$. It follows that $\left\vert{S}\right\vert = \mathfrak{c}$.

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Well, by Cantor's diagonal it can't be $\aleph_0$. (If it were, list all the sequences; change the i-th term of the i-th sequence; the resulting sequence converges to 1 but wasn't on the list.)

So the question is, is it going to be any cardinality between $\aleph_0$ and $\mathfrak{c}$.

hmm, don't think that's possible. But I'm weak on my theory. I'd think if $\{a_n\} \rightarrow r \in \mathbb R$ then $a_n/r \le b_n \le a_n/r + 1/n; b_n \in \mathbb Q$, then $\{b_n\} \rightarrow 1$. So I think it must be $\mathfrak{c}$.

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