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The definition of enumerability i have is "A set $\Bbb{X}$ is enumerable if, and only if, it has a bijection with the $\Bbb{N}$atural set ($f:\Bbb{N}\to\Bbb{X})$".

Well, as we know, a sequence is defined as a "function $g: \Bbb{N}\to\Bbb{R}$".

Now, consider a growing sequence $(a_n)_{n\in\Bbb{N}}$. We have two positions: * converges; * diverges. If it diverges, i don't know what to say, 'cause the sequence could either be $\{1,2,3,...\}$ as it could be $\{e,e^2,e^3,...\}$ (i need help here).

On the other hand, if it converges to $k$, we know we can find a $n_0\in\Bbb{N}$ such that for all $\epsilon>0$, we have $|x_n-k|<\epsilon=(k-\epsilonn_0$ .

The sequence is monotonic growing and converges, hence is limited by $k=\sup(x_n)$.

Considering now the set of finitely many terms of the sequence before de index $n_0$ as an enumerable set (cause it's finite), we must show that $\Bbb{Y}=\{x_n|k-\epsilon

Well, the convergent sequence is also a cauchy sequence, hence $\exists n_1; \forall\delta>0$ we have $|x_m-x_p|<\delta, \forall m,p>n_1$. This is the same as saying that two terms $x_m$ and $x_p$ is $\delta$ far from each other on a $\Bbb{R}$eal interval. As we also know, any $\Bbb{R}$eal interval is not enumerable, hence taking $\max\{\epsilon,\delta\}$, any interval there will not be enumerate. The union between enumerate and no enumerate sets is not enumerate, hence the set is not enumerate.

I made this proof, but i'm not so sure about it. And the part above about the not convergence is also making me mad! Any help is welcome, guys!

Thankyou!

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    If $(a_n)$ is really a sequence of _real_ numbers, it is clear that the set of such sequences is not enumerable, from only considering the first member $a_1$. The are uncountably many choices for $a_1$ alone, and two sequences are considered distinct provided that their first members are distinct. – 2017-01-07
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    I like your comment, man! 'Cause u'r saying that since a sequence is defined by $f:\Bbb{N}\to\Bbb{R}$, the image of course is uncontable. But why do we are considering just the growing sequence? Why the set of constant sequence is not enumerable, since the image isn't? And about the decreasing sequence's set? Thankyou! – 2017-01-08

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Notice that there is an injective map between the sequences in $\mathbb Z^+$ and the increasing sequences of $\mathbb N$:

$(a_1,a_2,a_3,a_4,\dots)\rightarrow (a_1,a_1+a_2,a_1+a_2+a_3,a_1+a_2+a_3+a_4\dots)$.

Since the first set is not countable, the second isn't either.

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    Why do the first set is not countable? – 2017-01-08
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    Because this set is clearly larger than $2^\mathbb N $ which is nota contable. – 2017-01-08
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    But, in this way, $\Bbb{Z}>\Bbb{Z}^+>2^{\Bbb{N}} $ and Z would not be enumerable. I think I got it wrong. – 2017-01-08
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    No, It is false that $Z> 2^\mathbb N$ – 2017-01-08
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    I got it now what you said, man. Thx! – 2017-01-08
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    No problem , my pleasure. – 2017-01-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51389/discussion-between-lucas-barbiere-and-jorge-fernandez-hidalgo). – 2017-01-08
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HINT: In fact you don’t need to deal with $\Bbb R$: the set of strictly increasing sequences of natural numbers is already not enumerable. Let $S$ be the set of sequences $\langle n_k:k\in\Bbb N\rangle$ such that $n_0=0$, and $n_{k+1}-n_k\in\{1,2\}$ for each $k\in\Bbb N$. For each $\sigma=\langle n_k:k\in\Bbb N\rangle\in S$ let

$$A(\sigma)=\{k\in\Bbb N:n_{k+1}-n_k=2\}\;.$$

Show that the function $\varphi:S\to\wp(\Bbb N):\sigma\mapsto A(\sigma)$ is a bijection. Then use the fact that $\wp(\Bbb N)$ is not enumerable to conclude that $S$ is not enumerable.

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    Hello, professor! I'm not understanding how does the function $\phi$ works. Hence, I dont know how to prove the bijection. I think I need a second hint, please. – 2017-01-08
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    @Lucas: If $\sigma=\langle 0,1,3,4,6,7,9,10,\ldots\rangle$, so that the differences $n_{k+1}-n_k$ are alternately $1,2,1,2,1,2,\ldots$, then $n_{k+1}-n_k=2$ if and only if $k$ is odd. Then $\varphi(\sigma)=A(\sigma)=\{1,3,5,7,\ldots\}$, the set of odd natural numbers. If $\sigma=\langle 0,2,3,5,6,8,9,\ldots\rangle$, so that $n_{k+1}-n_k=2$ if and only if $k$ is even, then $\varphi(\sigma)=A(\sigma)=\{0,2,4,6,8,\ldots\}$, the set of even natural numbers. If $B$ is any subset of $\Bbb N$, let $\chi_B$ be the characteristic function of $B$. Let $n_0=0$, and for $k\in\Bbb N$ let ... – 2017-01-08
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    ... $n_{k+1}=n_k+\chi_B(k)+1$. Let $\sigma=\langle n_k:k\in\Bbb N\rangle$; it’s not hard to check that $\varphi(\sigma)=B$. – 2017-01-08