Cardinality of the set of all functions from N to N

Flying_Goat
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Homework Statement


Let NN be the set of all functions from N to N. Prove that |NN|=c


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The Attempt at a Solution



I can prove that the set of all functions from N to {0,1} has cardinality of the continuum, but i can't generalise it. Any help would be appreciated.
 
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Let f be a function from N to N. Construct a number, x, by writing a decimal point, then 0.f(1)f(2)f(3)... so that each function is mapped to the number, between 0 and 1, having the values of f as digits. That maps each function to a number. What numbers?
 
Thanks, I had thought of your argument when I was trying to prove |P(N)|=c but it didn't work out...I can't believe that it works for this question lol. Anyway thanks for your help.
 
Is the map injective? Because I could have f(2n-1)=12,f(2n)=3 or f(2n-1)=1,f(2n)=23 and both of these functions would give me 0.123123...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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