Cardinality of the set of all roots of natural numbers

Question

I found this to be an interesting question.

Define $S$ = {$n^{1/k}|n, k \in \Bbb N$}.

Now clearly $\Bbb N \subset S$ $\to |S|\ge |\Bbb N|$.

Let $T$ = {$\frac{1}{k}$|$k\in \Bbb N$}.

Define $f:T \to \Bbb N$ by $f(x) = \frac{1}{x}$. Then |$T$| = |$\Bbb N$|.

Now we use the fact that |$A^{B}$| = |$A$|$^{|B|}$.

I.e., |$S$| = $\aleph_0^{\aleph_0}$.

Now, how do we go forward? [Note: $0$ is not included in $\Bbb N$ here].

Aleph-Null^Aleph_null is the same as 2^alephnull. This is the cardinality of the continuum. — Jacob Wakem, Nov 13 '16 at 18:46
Your $S$ is a subset of the set of algebraic numbers which is countable, see for example Proving the countability of algebraic numbers. — dxiv, Nov 13 '16 at 18:57

score 0 · Answer 1 · answered Nov 13 '16 at 18:49

0

Your argument is flawed. We are not choosing the functions from one set to the other but merely pairs including one from either set (and this overcounts). Thus you want alephnull x alephnull is alephnull. The overcounting doesn't hurt because it clearly has a subset of size alephnull.

answered Nov 13 '16 at 18:49

Jacob Wakem

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1

Relevant username :) – Wojowu Nov 13 '16 at 18:56
@wojowu that's what I was thinking. :D – Simply Beautiful Art Nov 13 '16 at 19:01

Cardinality of the set of all roots of natural numbers

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