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(This is similar to this question) I'm working in a proof of the following proposition: Consider the set of all polynomials with rational coefficients

$\mathbb{Q}[t]=\{a_0+a_1t+a_2t^2+...+a_nt^n:n\in\mathbb{N}\,and\,a_0,...,a_n\in\mathbb{Q}\}$

The set $\mathbb{Q}[t]$ is countable.

Proof. What uniquely identifies any polynomial (i.e. any element of $\mathbb{Q}[t]$) is the combination of coefficients that it has. Then I can define the set of all the n-degree polynomials represented by its coefficients

$A_n=\{(a_0,a_1,...,a_n):a_k\in\mathbb{Q}\;for\;0 \le k\le n\;and\;a_n\ne0\}$

Note that $A_n\subseteq\mathbb{Q}^n$ implying that $A_n$ is countable. So since $\mathbb{Q}[t]=\cup_{i=1}^{\infty}A_i$, therefore $\mathbb{Q}[t]$ must be countable. $\square$

Is this reasoning right?

Thank you.

manifold
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    The proof looks pretty good, though technically, it may not be correct to say the set of polynomials of degree $n$ is actually equal to $A_n$ - merely that we can produce a bijection between the two. Also, you forgot the zero polynomial, which is not in any $A_n$. – Mark Saving Jul 14 '22 at 03:23
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    Very close, but note $\mathbb Q[t]\ne \cup A_n$. Instead if we define $P_n = {a_0+a_1t + .... + a_nt^n| (a_0, .....,a_n) \in A_n}$ then $\mathbb Q[t] = \cup P_n$. Now all you need to do is show that $|A_n| = |P_n|$ (obvious one to one). Also you claim without proof that $A_n$ is countable (although $A_n = \mathbb Q^n$ and you should have a specific theorem about finite cross products). As long as you avoid the trap that you don't have an infinite cross product you should be fine. – fleablood Jul 14 '22 at 03:29
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    One more small note: $A_n\subseteq \Bbb Q^{n+1}$. – Milten Jul 14 '22 at 03:50

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