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I'm trying to find the cardinality of the set of all polynomials with coefficients in ℝ. What's wrong with the following proof:

Let $f$ be a function: $$f: \mathbb R[x] \to P(\mathbb R)$$ $$f(a_{0}+a_{1}x^{1}+...+a_{k}x^{k}) = \left \{ a_{0}, a_{1}, ..., a_{k} \right \}$$

For example: $$f(4.3x+2.5) = \left \{ 4.3,2.5 \right \}$$

f is obviously not injective, but is onto. Meaning that $$|\mathbb R[x]]| \geq \left | P(R) \right | $$

What am I missing?

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  • $\begingroup$ Why is it onto? Can polynomials have an infinite number of coefficients? $\endgroup$ – Olivier May 16 '17 at 12:08
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    $\begingroup$ It doesn't look onto to me. What is a polynomial mapping to $\{0,1,2,3,\ldots\}$? $\endgroup$ – Joppy May 16 '17 at 12:08
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    $\begingroup$ @Rizon - what about the rest of the natural numbers? I meant the (infinite) set of whole numbers $0,1,2,3,4,\ldots$. A power set of a set is all subsets, not just finite subsets. $\endgroup$ – Joppy May 16 '17 at 12:18
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    $\begingroup$ @Joppy, Oh! I admit I didn't know a subset can be infinite. Silly me :| $\endgroup$ – Rizon May 16 '17 at 12:26
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    $\begingroup$ You mean the cardinality of the set of all polynomials with real coefficients. The cardinality of a polynomial is nothing of interest. $\endgroup$ – martin.koeberl May 16 '17 at 12:53
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Notice that the cardinality of polynomials of degree $0$ (Only free coefficients) is $|\mathbb{R}| = \mathfrak{c}$ (We just map such polynomials to their free coefficients).

The cardinality of polynomials of degree $1$ is $|\mathbb{R} \times \mathbb{R}| = \mathfrak{c}$.

The cardinality of polynomials of degree $2$ is $|\mathbb{R} \times \mathbb{R} \times \mathbb{R}| = \mathfrak{c}$

...

The cardinality of polynomials of degree $n$ is $|\mathbb{R}^n| = \mathfrak{c}$

...

Your set is just the countable union of all these sets from above, and therefore its cardinality is also $\mathfrak{c}$. (See here: Cardinality of union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$)

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Ok so look at the interval $[-n,n]$.Total number of subsets of it is $c$...now.coefficients come from one of these subsets for some $n \in N$. So it is less than $c* \aleph $...which is still $c$.The reverse inclusion is trivial.So cardinality of total polynomial is $c$.

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