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I have this question in my lecture notes and I've been pondering of over this since the morning.

I wish to represent the coefficients of any n-th degree polynomial by a set. Since this set has all the rational numbers (It's a subset of the set of rational numbers), hence If I further create the set of all such sets of coefficients, I shall have the power set of Q (set of sets of all possible rational numbers).

From what I understand, By Cantor's Theorem, we can state that any power set of a set (P(A)) has a cardinality greater than that of the set itself (A). I know that the cardinality of Q is aleph-null.

Then, By Cantor's Theorem, I can state that the power set of Q has the cardinality aleph-one, i.e., the set of all the polynomials with rational coefficients must have the cardinality aleph-one. Which is contrary to what I had to prove.

Can anyone please guide me where I'm making the mistake? (I believe there's some mistake at the point where I inherently assume that the set consisting of polynomials coefficient would be finite; I'm still unsure of it.)

Thanks in advance!

(Sorry for the fact that I don't know how to use LaTeX, apologies for that.)

Andrés E. Caicedo
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Mayank Sharma
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    A polynomial has only finitely many nonzero coefficients. – Angina Seng Aug 27 '17 at 12:27
  • @LordSharktheUnknown So, the problem lies in the fact that I assumed the subset of Q to be finite, of whose collection can never be used to generate the power set of Q. I mean, however large may the number of non-zero rational coefficients be, it's still a finite number. Am I right? – Mayank Sharma Aug 27 '17 at 12:30
  • for MathJax help see: http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm#alphaList and then use delimiters of dollars signs on the set of instructions: like \sum_\limits{i=1}^\infty i which makes $\sum_\limits{i=1}^\infty i$ once you add delimiters. –  Aug 27 '17 at 12:36
  • Minor comment but: "$\aleph_1$" (the first uncountable cardinal) is not necessarily the same thing as "$2^{\aleph_0}$" (the cardinality of $\mathbb{R}$). We certainly have $2^{\aleph_0}\ge\aleph_1$, but it is possible that $2^{\aleph_0}>\aleph_1$; in fact, it is known that the standard axioms of set theory can neither prove nor refute this. By Cantor, $\mathcal{P}(\mathbb{Q})$ is uncountable, and it's not hard to show that it has cardinality $=2^{\aleph_0}$, but there's no reason to believe it's exactly $\aleph_1$. – Noah Schweber Aug 27 '17 at 16:01

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