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Let $\mathbb{C}[x]$ denote the set of all polynomials with complex coefficients. Determine the cardinality of $\mathbb{C}[x]$ using an expression built up from cardinal operations, 2 and $\omega$.

I would appreciate if someone could give me a hint, as I have no idea how to begin. Thanks!

  • You can begin by calculating the cardinality of $\Bbb C[x]$. –  Jun 13 '20 at 09:08
  • I have tried to do it, but I am not sure, could it be $\omega$? –  Jun 13 '20 at 09:46
  • Why would that be? –  Jun 13 '20 at 13:01
  • As $\mathbb{C}$ is equal to $|\mathbb{R} \times \mathbb{R}|$ which is then equal to $|\mathbb{R}|$, the cardinality of $\mathbb{C}$ is then equal to $2^{\aleph_0}$. By the continuum hypothesis the cardinality of $\omega$ is smaller than the cardinality of the real numbers. It was foolish of me to just name omega without thinking. –  Jun 13 '20 at 16:07
  • The continuum hypothesis is not a theorem. Perhaps you were referring to Cantor's theorem. –  Jun 13 '20 at 16:25

2 Answers2

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Sketch/hint: You seem to be aware of the fact that $\lvert \Bbb R\times\Bbb R\rvert=\lvert \Bbb R\rvert$. From this it follows that $\lvert \Bbb R^n\rvert=\lvert \Bbb R\rvert$ for all $n\ge 1$. Assuming for convenience that all the $\Bbb R^n$-s (or $\Bbb C^n$-s, it's the same) are disjoint, is it possible that $\left\lvert\bigcup_{n\ge1}\Bbb R^n\right\rvert=\lvert\Bbb R\rvert$ as well? Finally: can you find an injection $\Bbb C[x]\to\bigcup_{n\ge 1}\Bbb C^n$ ?

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The cardinality of $\mathbb{C}$ is $| \mathbb{C} | = |\mathbb{R} ^2| = (2^{\aleph_0})^2 = 2^{2\aleph_0} = 2^{\aleph_0}$.

This is based on a theorem saying that for any $\aleph_\alpha$ and $\aleph_\beta$, $\aleph_\alpha + \aleph_\beta = \aleph_{max(\alpha, \beta)}$ (the proof can for example be found in Thomas Jech's Set Theory).

The same way, the cardinality of $\mathbb{C}^n$ is, $\forall n \in \mathbb{N}\setminus\{0\}, |\mathbb{C}^n| = 2^{n\aleph_0} = 2^{\aleph_0}$.

The cardinal of $\mathbb{C}[x]$ is

$$ |\mathbb{C}[x]| = |\bigcup_{n \in \mathbb{N}\setminus\{0\}} \mathbb{C}^n| = \sum_{n \in \mathbb{N}\setminus\{0\}} 2^{\aleph_0} = \aleph_0 2^{\aleph_0} = 2^{\aleph_0} $$

The last equality is because, actually the result you can look up (e.g. in Thomas Jech) is very strong:

$$ \forall \kappa,\lambda\mathrm{~infinite~cardinals}, \kappa+\lambda = \kappa\cdot\lambda = max(\kappa, \lambda) $$

Note: to be rigorous, $\omega$ is used mainly to name the first infinite ordinal, whereas for the cardinal one prefers using the notation $\aleph_0$.

almaus
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