How can i find a bijection from N( natural numbers) to Q[X]

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Discussion Overview

The discussion revolves around finding a bijection between the set of natural numbers, N, and the set of polynomials with rational coefficients, Q[X]. Participants explore the concept of countability and the challenges in explicitly defining such a bijection.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asks for guidance on finding a bijection from N to Q[X], expressing difficulty in identifying a solution.
  • Another participant outlines a method to demonstrate that Q[X] is countable by constructing sets A_n for polynomials of increasing degree, concluding that the union of these countable sets is countable.
  • A later reply questions how to explicitly write down the bijection function from N to Q[X], noting that while a bijection exists, it may not be straightforward to express.
  • Another participant suggests that while it is challenging to explicitly define the bijection, the existence of such a function is often sufficient for most discussions.

Areas of Agreement / Disagreement

Participants generally agree that a bijection exists between N and Q[X], but there is no consensus on how to explicitly define this function. The difficulty in expressing the bijection is acknowledged by multiple participants.

Contextual Notes

The discussion highlights the complexity involved in explicitly constructing bijections, particularly between sets like N and Q[X], and the potential for these functions to be cumbersome or "ugly" in form.

cghost
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How can i find a bijection from N( natural numbers) to Q[X] ( polynomials with coefficient in rational numbers ). I can't find a solution for this. Can you please point me in the right direction ?
 
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This is a standard trick. The proof is as follows:

Let [tex]A_0=\mathbb{Q}[/tex]. This is clearly countable.
Let [tex]A_1=\{a+bx~\vert~a,b\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times \mathbb{Q}[/tex].
Let [tex]A_2=\{a+bx+cx^2~\vert~a,b,c\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}[/tex].

Continue with this process. Finally, we have [tex]\mathbb{Q}[x]=\bigcup_{n\in \mathbb{N}}{A_n}[/tex]. This is countable as countable union of countable sets.
 


micromass said:
This is a standard trick. The proof is as follows:

Let [tex]A_0=\mathbb{Q}[/tex]. This is clearly countable.
Let [tex]A_1=\{a+bx~\vert~a,b\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times \mathbb{Q}[/tex].
Let [tex]A_2=\{a+bx+cx^2~\vert~a,b,c\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}[/tex].

Continue with this process. Finally, we have [tex]\mathbb{Q}[x]=\bigcup_{n\in \mathbb{N}}{A_n}[/tex]. This is countable as countable union of countable sets.

You proved here that there is a bijection from N to Q[x], but how can i write that function down ?
Or you can't exactly explicit the function f : N -> Q[x] , f(x) = ... ?
If it's a bijection, every natural number should point to a single polynom, is that right?
Or Q[X] is defined as a set of equivalence classes.
 


It's very hard to explicitely say what the bijection is. Most of the time people are happy with knowing that there exists a bijection.

Of course, if you examine all the steps I made, maybe you can make the bijection step-by-step. But this will be a very ugly thing. In fact, even a bijection between [tex]\mathbb{N}[/tex] and [tex]\mathbb{Q}[/tex] is hard and ugly to write down...
 


Thanks for helping me
 

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