$X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$

Question

Let $X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$

Prove $$\vert X\vert = \vert \mathbb R\vert=c$$

without using $2^{\aleph_0} = \aleph_0^{\aleph_0} = c$

My Attempt

I know that $\vert X\vert = \vert \mathbb R\vert=c$, using this $$\forall n \geq 2, n^{\aleph_0} = \aleph_0^{\aleph_0} = c$$ (I still haven't proved that).
I want to show this in another way.
Probably looking for a surjective function between $X$ and $\mathbb{R}$ or the sucessions of natural numbers.

The thing is that the latter demonstration is after the exercise I'm asking about, so I believe that there is a way to prove it without using it, but I haven't found a way.

Is the objective of the task to find the cardinality of $X$? — ViktorStein, Jan 14 '20 at 15:48
I've just edited. Hope it's more clear, but I want to prove that the cardinality of $X$ is $c$ without using that $n^{\aleph_0} = c$ — Silkking, Jan 14 '20 at 15:59

score 1 · Answer 1 · answered Jan 15 '20 at 21:59

1

See that $|X|\leq c$ is easy. For the other inequality, you consider $f: [0,1]\to X$ defined by $f(0.a_1 a_2 a_3 ...) = (a_1, a_1 +a_2, a_1 +a_2 + a_3, ...)$ and you can check that $f$ is one-to-one (finally you use that $|[0,1]|=c$).

answered Jan 15 '20 at 21:59

juancodmw

547

Marvelous solution. Thank you. There is a tiny but easily solvable problem. The sucessions are natural numbers. Adding $1$ to $a_1$ solves it – Silkking Jan 16 '20 at 22:50
yes!, you must add 1 to all coordenates and solve it. – juancodmw Jan 17 '20 at 12:42

$X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$

My Attempt

1 Answers1