$\vert\mathbb{Q}^m\vert = \vert\mathbb{N}\vert$. Proof

Question

To start with I'm not completely understand what $\mathbb{Q}^m$ is. Is it the number of all elements in $\mathbb{Q}$ that was taken to some power $m$?

To show that two sets have the same cardinality we must introduce a map $f: \mathbb{N} \longrightarrow \mathbb{Q}^m$ that would be a bijection, but I got stuck with it. Can someone please show me a proof or give a hint?

The set $\mathbb Q^m$ is the set of all sequences $(q_1, \ldots, q_m)$ where each $q_i$ is a rational number. (You may be more familiar with the set $\mathbb R^m$ -- the idea is the same.) — Mees de Vries, Dec 12 '18 at 10:47

score 0 · Accepted Answer · answered Dec 12 '18 at 10:46

0

We have an explicit bijection between $\Bbb{N}$ and $\Bbb{Q}$, see

Produce an explicit bijection between rationals and naturals?,

and an bijection between $\Bbb{Q}$ and $\Bbb{Q}^m$, see

Does There exist a continuous bijection $\mathbb{Q}\to \mathbb{Q}\times \mathbb{Q}$?

Is $\mathbb Q \times \mathbb Q $ a denumerable set?

Now compose the two bijections.

answered Dec 12 '18 at 10:46

Dietrich Burde

130,978

If your answer is just links to other questions, then perhaps the question is just a duplicate, and instead of writing an answer you should close it as such? – Asaf Karagila Dec 12 '18 at 10:52
@AsafKaragila Yes, I agree. Which one of the three should I chose? – Dietrich Burde Dec 12 '18 at 12:08
Since we can have multiple duplicate targets, why not all of them? – Asaf Karagila Dec 12 '18 at 12:09
Ah, I didn't know that multiple assigments are possible, thank you! – Dietrich Burde Dec 12 '18 at 12:11

$\vert\mathbb{Q}^m\vert = \vert\mathbb{N}\vert$. Proof

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