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?
Best Answer
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 $|\mathbb{R} \times \mathbb{R} \times \mathbb{R}| = \mathfrak{c}$
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The cardinality of polynomials of degree $n$ is $|\mathbb{R}^n| = \mathfrak{c}$
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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 $\mathfrak{c}$)