Is $T$, the set of all polynomials with integer coefficients, countable?
Here is my attempt at a proof:
Let $S$ = {$ (a_0, a_1, …, a_k)$ | $a_i\in \Bbb Z$, $k \in \Bbb N$}
$T$ is the set in question. That is, $T$ = {$ (a_0, a_1x, …, a_kx^{k})$ | $a_i \in \Bbb Z$, $k \in \Bbb N$}.
Then by the enumeration principle $S$ is countable, i.e., |$S$| = |$\Bbb N$|, because it can be labelled by
$L =$ {$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, , , (, )$} [Note: I've included "," as an element].
Now define an injection $f: T \to S$ by:
$f(a_0 + a_1x + … +a_kx^{k}$) = ($a_0, a_1, …, a_k$).
Then $|T| \le |S|$.
We can define a similar injection $g: S \to T$, establishing |$S| \le |T$|. So |$T$| = |$S$| = |$\Bbb N$| which is what we wanted to prove.
Is this good? I'm pretty new to set theory so I want to make sure these proofs are valid.
Best Answer
Any polynomial $p(x)$ over $\Bbb Z$ can be written as $p(x)=a_0+a_1x+\ldots +a_nx^n,n\in \Bbb N$.
Since $a_i\in \Bbb Z$ ,and $\Bbb Z$ is countable so for each $a_i$ we have countable choices .
EDIT:By "choices" I mean number of options available for us to select the $a_i's$ from.Since the polynomial is over $\Bbb Z$ so we have the option of selecting the $a_i's$ from the set of integers.
And since countable union of countable sets is countable so the set of all polynomials over $\Bbb Z$ is countable also