Irrational numbers Cardinality.

cardinalsset-theorysolution-verification

The real numbers, $\mathbb{R}$, are uncountable and the rational numbers, $\mathbb{Q}$, are countable. We can write $\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})$. Since $\mathbb{Q}$ is countable and $\mathbb{R}$ is uncountable, $\mathbb{R} \setminus \mathbb{Q}$ must be uncountable. If $\mathbb{R} \setminus \mathbb{Q}$ were countable, the union $\mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})$ would be countable, which contradicts the fact that $\mathbb{R}$ is uncountable. Therefore, $\mathbb{R} \setminus \mathbb{Q}$ must be uncountable.

The rational numbers, $\mathbb{Q}$, and the irrational numbers, $\mathbb{R} \backslash \mathbb{Q}$, are two disjoint sets that together form the real numbers. The cardinality of the union of two disjoint sets is given by the sum of the cardinalities of the individual sets. Therefore, we have:

$$|\mathbb{R}| = |\mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})| = |\mathbb{Q}| + |\mathbb{R} \setminus \mathbb{Q}|$$

The cardinality of $\mathbb{R}$ and $\mathbb{Q}$ is given by $|\mathbb{R}| = \mathfrak{c}$ and $|\mathbb{N}|=\aleph_0$. Now we use the theorem about cardinal addition to get:

$$\mathfrak{c} = \aleph_0 + |\mathbb{R} \setminus \mathbb{Q}| = |\mathbb{R} \setminus \mathbb{Q}|$$

Therefore, the cardinality of $\mathbb{R} \setminus \mathbb{Q}$ is $\mathfrak{c}$. Is this a correct proof or am I missing something?

Best Answer

The theorem in your comment is indeed reasonable, in fact it is true (with appropriate axioms, as suggested by J.G.).

But what you wrote in your comment under the main post sounds like this:

  • You have guessed a reasonable sounding generalization of what you have to prove;
  • You then based your proof on that generalization;
  • But you have not yet checked whether that generalization is true.

I would say that this is a version of circular proof... unless and until you understand a proof of that generalization.

But then I would also say that proving the generalization is harder than directly solving the problem you have in your hand (again as hinted in the comment of @J.G.).

There's a bunch of possible ways to solve your problem, here's one.

Choose a bijection $f : \mathbb N \mapsto \mathbb Q$. Choose an injection $F : \mathbb N \times \mathbb N \to \mathbb R$ having the property that $F(n,1)=f(n)$. For example, let $(p_m)_{m \in \mathbb N}$ be the sequence of primes with $1$ stuck on at the beginning, namely $(p_m) = (1,2,3,5,7,11,13,\ldots)$. We can define $$F(n,m) = f(n) \cdot \sqrt{p_m} $$ The fact that $F$ is injective takes a tiny bit of number theory to prove.

Finally, define a bijection $G : \mathbb R \to \mathbb R - \mathbb Q$ by the formula $$G(x) = \begin{cases} x & \quad\text{if $x \ne F(n,m)$ for all $n,m \in \mathbb N \times \mathbb N$} \\ F(n,m+1) &\quad\text{if $x=F(n,m)$} \end{cases} $$

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