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I want to calculate the cardinalities of the following sets:

$A = ${a ∈ $ {\mathbb {R}}^{{+}} | a^4∈{\mathbb {N}} $}

I belive it's $\!\, \aleph_0 $, but not sure how to prove it.

$B = ${a ∈ $ {\mathbb {R}}^{{+}} | ∃n∈{\mathbb {N}} ,$ $ a^n∈{\mathbb {N}} $}

I belive it's $\!\, \aleph $, but not sure how to prove it.

Any ideas?

2 Answers 2

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For the first one. The mapping $f: A \to \mathbb{N}$, $f(a) = a^4$ is a bijection from $A$ to $N$ and hence cardinality of $A$ is $\aleph_0$

For the next one. (As suggested by my friend) Observe that $\mathbb{N} \subseteq B$ and $B \subseteq \mathbb{A}$ where $\mathbb{A}$ denotes the set of algebraic numbers. Now, if you assume for the moment that $\mathbb{A}$ is countable, then it follows that $|B| = \aleph_0$.

Claim : $\mathbb{A}$ is countable

Proof : As $\mathbb{A}$ contains $\mathbb{Q}$, $\mathbb{A}$ is at least countable.

Let $S_n$ denote the set of algebraic numbers with minimal polynomial of degree $n$.

Note that $$\mathbb{A} = \bigcup_{n \geq 1} S_n$$

Hence, we'll be done if we show that $S_n$ is at most countable.

To do this, note that the minimal polynomial of degree $n$ can be encoded as a $n$ tuple of rational numbers. That is, $|S_n| \leq |\mathbb{Q}^n| = \aleph_0$. This completes the proof.

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Both sets are infinite countable. In the second case, we have the set of all $n$ th roots of the natural numbers. This set must be countable. The first is clear because we can list all the roots.

So, the answer is $\aleph_0$ in both cases.

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    Could you please provide further explanation about the second set? Also, could you please give me a lead how to prove the first is countable? – 2017-02-11
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    Consider the map sending a pair $(a,n)$ to the set of the (unique) $n$-th root of $a$. – 2017-02-11
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    Oh I think I see it now. And then I get a a countable union of finite sets? – 2017-02-11
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    But do not use $(a,n)$ (unlucky notation!) because $a$ stands for the real number, take another letter – 2017-02-11