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quarkine
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While studying the countability of sets, i came across the following problem (In Methods of Real Analysis, Goldberg) :
Show that Pn=set of polynomials of degree n (with all coefficients being integers and n fixed positive integer) is countable..
 one year ago
 one year ago
quarkine Group Title
While studying the countability of sets, i came across the following problem (In Methods of Real Analysis, Goldberg) : Show that Pn=set of polynomials of degree n (with all coefficients being integers and n fixed positive integer) is countable..
 one year ago
 one year ago

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quarkine Group TitleBest ResponseYou've already chosen the best response.0
my try was
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.2
All you need is that \(\mathbb{Z}^n\) is countable because from there you can simply setup a bijection between \(\mathbb{Z}^n\) and \(P_n\) with integer coefficients .
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.2
This is indeed by induction because \(Z \times Z\) is countable via cantor's pairing function. The same technique used to show that \(\mathbb{Q}\) is countable. Then do the same thing inductively \(\mathbb{Z}^2 \times Z\), etc.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.2
Sorry I meant \(\mathbb{Z}^{n + 1}\)
 one year ago

quarkine Group TitleBest ResponseYou've already chosen the best response.0
Hmm i guess that's true but i just needed to know if the method i mentioned above will work.I found it wont :(
 one year ago

quarkine Group TitleBest ResponseYou've already chosen the best response.0
Thanks for the answer anyway..
 one year ago

kbomeisl Group TitleBest ResponseYou've already chosen the best response.1
Alchemista is largely correct in that setting up a function between the two given sets is the most parsimonious way to go about such a proof. However, you don't really need a bijection, in fact, a surjection with the natural numbers as the image or an injection with the natural numbers as the preimage would be sufficient. I'm having some trouble interpreting your proof, from what I see, you are mapping a ntuple of integer coefficients to one natural number. But I can't quite understand what method you are using to map. Honestly, this method will only work for finite term polynomials (which i know is the objective of the assignment) but a mathematician must always work for maximum generalizability. If you start working with arbitrarily number term polynomials, this proof method will fail as Cantor's diagonal method can be used to prove that indeed the cardinality of such polynomials is uncountable. When finding a surjection between a given set and the natural numbers, the function f(x,y,z..) = 2^x(3^y)(5^z)... with each base a prime and each power a member of the set. The fundamental theorem of arithmetic will guarentee that this is a bijection (think unique prime factoring) or course if one is using a general countable set that is not the natural numbers a unique prime factorization is not guarenteed so stick with the natural numbers.
 11 months ago

quarkine Group TitleBest ResponseYou've already chosen the best response.0
Thanks. What i wanted to do was to assign a unique no. with each polynomial.Your method seems great for that.Never occurred to me to use the fundamental theorem of arithmetic to generate unique number!
 11 months ago
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